I haven't found properly way how to decode base64URL to base64 format in swift.
According to base64url to base64 hJQWHABDBjoPHorYF5xghQ(base64URL) should be hJQWHABDBjoPHorYF5xghQ==(base64). Here could be more differences.
There are no solutions on stackoverflow.
"base64url" differs from the standard Base64 encoding in two aspects:
different characters are used for index 62 and 63 (- and _ instead
of + and /)
no mandatory padding with = characters to make the string length
a multiple of four.
(compare https://en.wikipedia.org/wiki/Base64#Variants_summary_table).
Here is a possible conversion function:
func base64urlToBase64(base64url: String) -> String {
var base64 = base64url
.replacingOccurrences(of: "-", with: "+")
.replacingOccurrences(of: "_", with: "/")
if base64.characters.count % 4 != 0 {
base64.append(String(repeating: "=", count: 4 - base64.characters.count % 4))
}
return base64
}
Example:
let base64url = "hJQWHABDBjoPHorYF5xghQ"
let base64 = base64urlToBase64(base64url: base64url)
print(base64) // hJQWHABDBjoPHorYF5xghQ==
if let data = Data(base64Encoded: base64) {
print(data as NSData) // <8494161c 0043063a 0f1e8ad8 179c6085>
}
For the sake of completeness, this would be the opposite conversion:
func base64ToBase64url(base64: String) -> String {
let base64url = base64
.replacingOccurrences(of: "+", with: "-")
.replacingOccurrences(of: "/", with: "_")
.replacingOccurrences(of: "=", with: "")
return base64url
}
Update for Swift 4:
func base64urlToBase64(base64url: String) -> String {
var base64 = base64url
.replacingOccurrences(of: "-", with: "+")
.replacingOccurrences(of: "_", with: "/")
if base64.count % 4 != 0 {
base64.append(String(repeating: "=", count: 4 - base64.count % 4))
}
return base64
}
Here is a cleaned up version of what Martin posted within a Swift 4 extension:
import Foundation
/// Extension for making base64 representations of `Data` safe for
/// transmitting via URL query parameters
extension Data {
/// Instantiates data by decoding a base64url string into base64
///
/// - Parameter string: A base64url encoded string
init?(base64URLEncoded string: String) {
self.init(base64Encoded: string.toggleBase64URLSafe(on: false))
}
/// Encodes the string into a base64url safe representation
///
/// - Returns: A string that is base64 encoded but made safe for passing
/// in as a query parameter into a URL string
func base64URLEncodedString() -> String {
return self.base64EncodedString().toggleBase64URLSafe(on: true)
}
}
extension String {
/// Encodes or decodes into a base64url safe representation
///
/// - Parameter on: Whether or not the string should be made safe for URL strings
/// - Returns: if `on`, then a base64url string; if `off` then a base64 string
func toggleBase64URLSafe(on: Bool) -> String {
if on {
// Make base64 string safe for passing into URL query params
let base64url = self.replacingOccurrences(of: "/", with: "_")
.replacingOccurrences(of: "+", with: "-")
.replacingOccurrences(of: "=", with: "")
return base64url
} else {
// Return to base64 encoding
var base64 = self.replacingOccurrences(of: "_", with: "/")
.replacingOccurrences(of: "-", with: "+")
// Add any necessary padding with `=`
if base64.count % 4 != 0 {
base64.append(String(repeating: "=", count: 4 - base64.count % 4))
}
return base64
}
}
}
This is objective c version of Base64URL to Base64
-(NSString *) base64urlToBase64 :(NSString *)base64url
{
NSString * base64 = [base64url stringByReplacingOccurrencesOfString:#"-" withString:#"+"];
base64 = [base64 stringByReplacingOccurrencesOfString:#"_" withString:#"/"];
if(base64.length % 4 != 0)
{
base64 = [base64 stringByAppendingString:[self stringWithRepeatString:"=" times:(4 - base64.length % 4)]];
}
return base64;
}
-(NSString*)stringWithRepeatString:(char*)characters times:(unsigned int)repetitions;
{
unsigned long stringLength = strlen(characters);
unsigned long repeatStringLength = stringLength * repetitions + 1;
char repeatString[repeatStringLength];
for (unsigned int i = 0; i < repetitions; i++) {
unsigned int pointerPosition = i * repetitions;
memcpy(repeatString + pointerPosition, characters, stringLength);
}
repeatString[repeatStringLength - 1] = 0;
return [NSString stringWithCString:repeatString encoding:NSASCIIStringEncoding];
}
Related
What's the most concise way to encode URL in Swift?
I have
let request_url = URL(string:"http://api.openweathermap.org/data/2.5/weather?appid=dca0aa44807a0bc05ed51c6a85472341&q="+"New York")
How can i add + instead of a space between New and York through some simple method available in swift?
If I understand you correctly, you want to encode spaces as + and encode other special characters correctly as well.
So this approach uses the native function addingPercentEncoding(withAllowedCharacters:) that translates a sspaces to %20. Afterwards we replace all occurrences of %20 with a +.
extension URL {
init?(encoding string: String) {
let encodedString = string
.addingPercentEncoding(withAllowedCharacters: .urlHostAllowed)
.replacingOccurrences(of: "%20", with: "+")
guard let encodedString != nil else { return nil }
self.init(string: encodedString!)
}
}
This is encapsulated in an URL init function, so you can use it like this:
let requestUrl = URL(encoding: "http://api.openweathermap.org/data/2.5/weather?appid=dca0aa44807a0bc05ed51c6a85472341&q="+"New York")
I want to replace all the non-alphanumeric characters (including white spaces) of a string up to the first alphabet
"\r\n A Simple PDF File \r\n This is a small demonstration .pdf file - \r\n just for use in the Virtual Mechanics tutorials. More text. And more \r\n text."
I used
let string = txtView.text
let newString = string.replacingOccurrences(of: "\n", with: "")
but it replaces all the "\n" of string. how could we replace the starting characters only?
I am able to remove leading white spaces using regular expression. This is the code :
extension String {
func trimLeadingWhitespaces() -> String {
return self.replacingOccurrences(of: "^\\s+", with: "", options: .regularExpression)
}
}
let string = " Some string abc"
let trimmed = string.trimLeadingWhitespaces()
print(trimmed)
There is a function drop that removes characters while a condition is filled.
Here are two variants of using it
let newString = string.drop { $0.isWhitespace || $0.isPunctuation }
let newString = string.drop { !($0.isLetter || $0.isNumber) }
You can use trimmingCharacters method of string class. It will remove all the leading and trailing whitespaces and \n from string.
let str = "\r\n A Simple PDF File \r\n This is a small demonstration .pdf file - \r\n just for use in the Virtual Mechanics tutorials. More text. And more \r\n text."
let result = str.trimmingCharacters(in: .whitespacesAndNewlines)
print(result)
Only leading spaces:
extension String {
func removingLeadingSpaces() -> String {
guard let index = firstIndex(where: { !CharacterSet(charactersIn: String($0)).isSubset(of: .whitespaces) }) else {
return self
}
return String(self[index...])
}
}
you can use two func from extension (thx #amer):
extension String {
func replacingFirstOccurrence(of string: String, with replacement: String) -> String {
guard let range = self.range(of: string) else { return self }
return replacingCharacters(in: range, with: replacement)
}
func removingLeadingSpaces() -> String {
guard let index = firstIndex(where: { !CharacterSet(charactersIn: String($0)).isSubset(of: .whitespaces) }) else {
return self
}
return String(self[index...])
}
}
and use it:
let myString = "\r\n A Simple PDF File \r\n This is a small demonstration .pdf file - \r\n just for use in the Virtual Mechanics tutorials. More text. And more \r\n text."
let charactersRemoved = myString.replacingFirstOccurrence(of: "\r\n", with: "")
let whiteSpacesRemoved = charactersRemoved.removingLeadingSpaces()
print(whiteSpacesRemoved) // "A Simple PDF File \r\n This is a small demonstration .pdf file - \r\n just for use in the Virtual Mechanics tutorials. More text. And more \r\n text."
[ |\t|\n|\r] matchs for space, newline, tabulation or carriage return
{2,} "2 times or more"
each match is replaced with a newline
let str = "\r\n A Simple PDF File \r\n This is a small demonstration .pdf file - \r\n just for use in the Virtual Mechanics tutorials. More text. And more \r\n text."
let newString = str.replacingOccurrences(
of: "[ |\t|\n|\r]{2,}",
with: "\n",
options: .regularExpression
)
print(newString)
// A Simple PDF File
// This is a small demonstration .pdf file -
// just for use in the Virtual Mechanics tutorials. More text. And more
// text.
As per your problem statement you need to delete non-alphabetic char which includes spaces, line,'#','$','%','^', etc. There is much non-alphabetic char. Instead of checking non-alphabetic char , you chan validate alphabetic char alone. Please check the following solution to your problem.
extension String{
func findFirstAlphabetic() -> String.Index?{
for index in self.indices{
if String(self[index]).isAlphanumeric == true{
return index
}
}
return nil
}
var isAlphanumeric: Bool {
return !isEmpty && range(of: "[^a-zA-Z0-9]", options: .regularExpression) == nil
}
func alphabetic_Leading_SubString() -> String?{
if let startIndex = self.findFirstAlphabetic(){
let newSubString = self[startIndex..<self.endIndex]
return String(newSubString)
}
return nil
}
}
usage:-
print("Output" + string.alphabetic_Leading_SubString())
Input :-
let string = "\r\n# # $ % & ^ * () -+ ######## A Simple PDF File \r\n This is a small demonstration .pdf file - \r\n just for use in the Virtual Mechanics tutorials. More text. And more \r\n text."
Output:-
If I encode a string like this:
var escapedString = originalString.stringByAddingPercentEscapesUsingEncoding(NSUTF8StringEncoding)
it doesn't escape the slashes /.
I've searched and found this Objective C code:
NSString *encodedString = (NSString *)CFURLCreateStringByAddingPercentEscapes(
NULL,
(CFStringRef)unencodedString,
NULL,
(CFStringRef)#"!*'();:#&=+$,/?%#[]",
kCFStringEncodingUTF8 );
Is there an easier way to encode an URL and if not, how do I write this in Swift?
Swift 3
In Swift 3 there is addingPercentEncoding
let originalString = "test/test"
let escapedString = originalString.addingPercentEncoding(withAllowedCharacters: .urlHostAllowed)
print(escapedString!)
Output:
test%2Ftest
Swift 1
In iOS 7 and above there is stringByAddingPercentEncodingWithAllowedCharacters
var originalString = "test/test"
var escapedString = originalString.stringByAddingPercentEncodingWithAllowedCharacters(.URLHostAllowedCharacterSet())
println("escapedString: \(escapedString)")
Output:
test%2Ftest
The following are useful (inverted) character sets:
URLFragmentAllowedCharacterSet "#%<>[\]^`{|}
URLHostAllowedCharacterSet "#%/<>?#\^`{|}
URLPasswordAllowedCharacterSet "#%/:<>?#[\]^`{|}
URLPathAllowedCharacterSet "#%;<>?[\]^`{|}
URLQueryAllowedCharacterSet "#%<>[\]^`{|}
URLUserAllowedCharacterSet "#%/:<>?#[\]^`
If you want a different set of characters to be escaped create a set:
Example with added "=" character:
var originalString = "test/test=42"
var customAllowedSet = NSCharacterSet(charactersInString:"=\"#%/<>?#\\^`{|}").invertedSet
var escapedString = originalString.stringByAddingPercentEncodingWithAllowedCharacters(customAllowedSet)
println("escapedString: \(escapedString)")
Output:
test%2Ftest%3D42
Example to verify ascii characters not in the set:
func printCharactersInSet(set: NSCharacterSet) {
var characters = ""
let iSet = set.invertedSet
for i: UInt32 in 32..<127 {
let c = Character(UnicodeScalar(i))
if iSet.longCharacterIsMember(i) {
characters = characters + String(c)
}
}
print("characters not in set: \'\(characters)\'")
}
You can use URLComponents to avoid having to manually percent encode your query string:
let scheme = "https"
let host = "www.google.com"
let path = "/search"
let queryItem = URLQueryItem(name: "q", value: "Formula One")
var urlComponents = URLComponents()
urlComponents.scheme = scheme
urlComponents.host = host
urlComponents.path = path
urlComponents.queryItems = [queryItem]
if let url = urlComponents.url {
print(url) // "https://www.google.com/search?q=Formula%20One"
}
extension URLComponents {
init(scheme: String = "https",
host: String = "www.google.com",
path: String = "/search",
queryItems: [URLQueryItem]) {
self.init()
self.scheme = scheme
self.host = host
self.path = path
self.queryItems = queryItems
}
}
let query = "Formula One"
if let url = URLComponents(queryItems: [URLQueryItem(name: "q", value: query)]).url {
print(url) // https://www.google.com/search?q=Formula%20One
}
Swift 4 & 5
To encode a parameter in URL I find using .alphanumerics character set the easiest option:
let urlEncoded = value.addingPercentEncoding(withAllowedCharacters: .alphanumerics)
let url = "http://www.example.com/?name=\(urlEncoded!)"
Using any of the standard Character Sets for URL Encoding (like .urlQueryAllowed or .urlHostAllowed) won't work, because they do not exclude = or & characters.
Note that by using .alphanumerics it will encode some characters that do not need to be encoded (like -, ., _ or ~ -– see 2.3. Unreserved characters in RFC 3986). I find using .alphanumerics simpler than constructing a custom character set and do not mind some additional characters to be encoded. If that bothers you, construct a custom character set as is described in How to percent encode a URL String, like for example:
// Store allowed character set for reuse (computed lazily).
private let urlAllowed: CharacterSet =
.alphanumerics.union(.init(charactersIn: "-._~")) // as per RFC 3986
extension String {
var urlEncoded: String? {
return addingPercentEncoding(withAllowedCharacters: urlAllowed)
}
}
let url = "http://www.example.com/?name=\(value.urlEncoded!)"
Warning: The urlEncoded parameter is force unwrapped. For invalid unicode string it might crash. See Why is the return value of String.addingPercentEncoding() optional?. Instead of force unwrapping urlEncoded! you can use urlEncoded ?? "" or if let urlEncoded = urlEncoded { ... }.
Swift 5:
extension String {
var urlEncoded: String? {
let allowedCharacterSet = CharacterSet.alphanumerics.union(CharacterSet(charactersIn: "~-_."))
return self.addingPercentEncoding(withAllowedCharacters: allowedCharacterSet)
}
}
print("\u{48}ello\u{9}world\u{7}\u{0}".urlEncoded!) // prints Hello%09world%07%00
print("The string ü#foo-bar".urlEncoded!) // prints The%20string%20%C3%BC%40foo-bar
Swift 3:
let originalString = "http://www.ihtc.cc?name=htc&title=iOS开发工程师"
1. encodingQuery:
let escapedString = originalString.addingPercentEncoding(withAllowedCharacters:NSCharacterSet.urlQueryAllowed)
result:
"http://www.ihtc.cc?name=htc&title=iOS%E5%BC%80%E5%8F%91%E5%B7%A5%E7%A8%8B%E5%B8%88"
2. encodingURL:
let escapedString = originalString.addingPercentEncoding(withAllowedCharacters: .urlHostAllowed)
result:
"http:%2F%2Fwww.ihtc.cc%3Fname=htc&title=iOS%E5%BC%80%E5%8F%91%E5%B7%A5%E7%A8%8B%E5%B8%88"
Swift 4 & 5 (Thanks #sumizome for suggestion. Thanks #FD_ and #derickito for testing)
var allowedQueryParamAndKey = NSCharacterSet.urlQueryAllowed
allowedQueryParamAndKey.remove(charactersIn: ";/?:#&=+$, ")
paramOrKey.addingPercentEncoding(withAllowedCharacters: allowedQueryParamAndKey)
Swift 3
let allowedQueryParamAndKey = NSCharacterSet.urlQueryAllowed.remove(charactersIn: ";/?:#&=+$, ")
paramOrKey.addingPercentEncoding(withAllowedCharacters: allowedQueryParamAndKey)
Swift 2.2 (Borrowing from Zaph's and correcting for url query key and parameter values)
var allowedQueryParamAndKey = NSCharacterSet(charactersInString: ";/?:#&=+$, ").invertedSet
paramOrKey.stringByAddingPercentEncodingWithAllowedCharacters(allowedQueryParamAndKey)
Example:
let paramOrKey = "https://some.website.com/path/to/page.srf?a=1&b=2#top"
paramOrKey.addingPercentEncoding(withAllowedCharacters: allowedQueryParamAndKey)
// produces:
"https%3A%2F%2Fsome.website.com%2Fpath%2Fto%2Fpage.srf%3Fa%3D1%26b%3D2%23top"
This is a shorter version of Bryan Chen's answer. I'd guess that urlQueryAllowed is allowing the control characters through which is fine unless they form part of the key or value in your query string at which point they need to be escaped.
Swift 4:
It depends by the encoding rules followed by your server.
Apple offer this class method, but it don't report wich kind of RCF protocol it follows.
var escapedString = originalString.addingPercentEncoding(withAllowedCharacters: .urlHostAllowed)!
Following this useful tool you should guarantee the encoding of these chars for your parameters:
$ (Dollar Sign) becomes %24
& (Ampersand) becomes %26
+ (Plus) becomes %2B
, (Comma) becomes %2C
: (Colon) becomes %3A
; (Semi-Colon) becomes %3B
= (Equals) becomes %3D
? (Question Mark) becomes %3F
# (Commercial A / At) becomes %40
In other words, speaking about URL encoding, you should following the RFC 1738 protocol.
And Swift don't cover the encoding of the + char for example, but it works well with these three # : ? chars.
So, to correctly encoding each your parameter , the .urlHostAllowed option is not enough, you should add also the special chars as for example:
encodedParameter = parameter.replacingOccurrences(of: "+", with: "%2B")
Hope this helps someone who become crazy to search these informations.
Everything is same
var str = CFURLCreateStringByAddingPercentEscapes(
nil,
"test/test",
nil,
"!*'();:#&=+$,/?%#[]",
CFStringBuiltInEncodings.UTF8.rawValue
)
// test%2Ftest
Swift 4.2
A quick one line solution. Replace originalString with the String you want to encode.
var encodedString = originalString.addingPercentEncoding(withAllowedCharacters: CharacterSet(charactersIn: "!*'();:#&=+$,/?%#[]{} ").inverted)
Online Playground Demo
This is working for me in Swift 5. The usage case is taking a URL from the clipboard or similar which may already have escaped characters but which also contains Unicode characters which could cause URLComponents or URL(string:) to fail.
First, create a character set that includes all URL-legal characters:
extension CharacterSet {
/// Characters valid in at least one part of a URL.
///
/// These characters are not allowed in ALL parts of a URL; each part has different requirements. This set is useful for checking for Unicode characters that need to be percent encoded before performing a validity check on individual URL components.
static var urlAllowedCharacters: CharacterSet {
// Start by including hash, which isn't in any set
var characters = CharacterSet(charactersIn: "#")
// All URL-legal characters
characters.formUnion(.urlUserAllowed)
characters.formUnion(.urlPasswordAllowed)
characters.formUnion(.urlHostAllowed)
characters.formUnion(.urlPathAllowed)
characters.formUnion(.urlQueryAllowed)
characters.formUnion(.urlFragmentAllowed)
return characters
}
}
Next, extend String with a method to encode URLs:
extension String {
/// Converts a string to a percent-encoded URL, including Unicode characters.
///
/// - Returns: An encoded URL if all steps succeed, otherwise nil.
func encodedUrl() -> URL? {
// Remove preexisting encoding,
guard let decodedString = self.removingPercentEncoding,
// encode any Unicode characters so URLComponents doesn't choke,
let unicodeEncodedString = decodedString.addingPercentEncoding(withAllowedCharacters: .urlAllowedCharacters),
// break into components to use proper encoding for each part,
let components = URLComponents(string: unicodeEncodedString),
// and reencode, to revert decoding while encoding missed characters.
let percentEncodedUrl = components.url else {
// Encoding failed
return nil
}
return percentEncodedUrl
}
}
Which can be tested like:
let urlText = "https://www.example.com/폴더/search?q=123&foo=bar&multi=eggs+and+ham&hangul=한글&spaced=lovely%20spam&illegal=<>#top"
let url = encodedUrl(from: urlText)
Value of url at the end: https://www.example.com/%ED%8F%B4%EB%8D%94/search?q=123&foo=bar&multi=eggs+and+ham&hangul=%ED%95%9C%EA%B8%80&spaced=lovely%20spam&illegal=%3C%3E#top
Note that both %20 and + spacing are preserved, Unicode characters are encoded, the %20 in the original urlText is not double encoded, and the anchor (fragment, or #) remains.
Edit: Now checking for validity of each component.
For Swift 5 to endcode string
func escape(string: String) -> String {
let allowedCharacters = string.addingPercentEncoding(withAllowedCharacters: CharacterSet(charactersIn: ":=\"#%/<>?#\\^`{|}").inverted) ?? ""
return allowedCharacters
}
How to use ?
let strEncoded = self.escape(string: "http://www.edamam.com/ontologies/edamam.owl#recipe_e2a1b9bf2d996cbd9875b80612ed9aa4")
print("escapedString: \(strEncoded)")
Had need of this myself, so I wrote a String extension that both allows for URLEncoding strings, as well as the more common end goal, converting a parameter dictionary into "GET" style URL Parameters:
extension String {
func URLEncodedString() -> String? {
var escapedString = self.addingPercentEncoding(withAllowedCharacters: .urlHostAllowed)
return escapedString
}
static func queryStringFromParameters(parameters: Dictionary<String,String>) -> String? {
if (parameters.count == 0)
{
return nil
}
var queryString : String? = nil
for (key, value) in parameters {
if let encodedKey = key.URLEncodedString() {
if let encodedValue = value.URLEncodedString() {
if queryString == nil
{
queryString = "?"
}
else
{
queryString! += "&"
}
queryString! += encodedKey + "=" + encodedValue
}
}
}
return queryString
}
}
Enjoy!
This one is working for me.
func stringByAddingPercentEncodingForFormData(plusForSpace: Bool=false) -> String? {
let unreserved = "*-._"
let allowed = NSMutableCharacterSet.alphanumericCharacterSet()
allowed.addCharactersInString(unreserved)
if plusForSpace {
allowed.addCharactersInString(" ")
}
var encoded = stringByAddingPercentEncodingWithAllowedCharacters(allowed)
if plusForSpace {
encoded = encoded?.stringByReplacingOccurrencesOfString(" ", withString: "+")
}
return encoded
}
I found above function from this link: http://useyourloaf.com/blog/how-to-percent-encode-a-url-string/.
let Url = URL(string: urlString.addingPercentEncoding(withAllowedCharacters: .urlQueryAllowed) ?? "")
None of these answers worked for me. Our app was crashing when a url contained non-English characters.
let unreserved = "-._~/?%$!:"
let allowed = NSMutableCharacterSet.alphanumeric()
allowed.addCharacters(in: unreserved)
let escapedString = urlString.addingPercentEncoding(withAllowedCharacters: allowed as CharacterSet)
Depending on the parameters of what you are trying to do, you may want to just create your own character set. The above allows for english characters, and -._~/?%$!:
What helped me was that I created a separate NSCharacterSet and used it on an UTF-8 encoded string i.e. textToEncode to generate the required result:
var queryCharSet = NSCharacterSet.urlQueryAllowed
queryCharSet.remove(charactersIn: "+&?,:;#+=$*()")
let utfedCharacterSet = String(utf8String: textToEncode.cString(using: .utf8)!)!
let encodedStr = utfedCharacterSet.addingPercentEncoding(withAllowedCharacters: queryCharSet)!
let paramUrl = "https://api.abc.eu/api/search?device=true&query=\(escapedStr)"
SWIFT 4.2
Sometimes this happened just because there is space in slug OR absence of URL encoding for parameters passing through API URL.
let myString = self.slugValue
let csCopy = CharacterSet(bitmapRepresentation: CharacterSet.urlPathAllowed.bitmapRepresentation)
let escapedString = myString!.addingPercentEncoding(withAllowedCharacters: csCopy)!
//always "info:hello%20world"
print(escapedString)
NOTE : Don't forget to explore about bitmapRepresentation.
version:Swift 5
// space convert to +
let mstring = string.replacingOccurrences(of: " ", with: "+")
// remove special character
var allowedQueryParamAndKey = NSCharacterSet.urlQueryAllowed
allowedQueryParamAndKey.remove(charactersIn: "!*'\"();:#&=+$,/?%#[]%")
return mstring.addingPercentEncoding(withAllowedCharacters: allowedQueryParamAndKey) ?? mstring
Swift 5
You can try .afURLQueryAllowed option if you want to encode string like below
let testString = "6hAD9/RjY+SnGm&B"
let escodedString = testString.addingPercentEncoding(withAllowedCharacters: .afURLQueryAllowed)
print(escodedString!)
//encoded string will be like en6hAD9%2FRjY%2BSnGm%26B
If I encode a string like this:
var escapedString = originalString.stringByAddingPercentEscapesUsingEncoding(NSUTF8StringEncoding)
it doesn't escape the slashes /.
I've searched and found this Objective C code:
NSString *encodedString = (NSString *)CFURLCreateStringByAddingPercentEscapes(
NULL,
(CFStringRef)unencodedString,
NULL,
(CFStringRef)#"!*'();:#&=+$,/?%#[]",
kCFStringEncodingUTF8 );
Is there an easier way to encode an URL and if not, how do I write this in Swift?
Swift 3
In Swift 3 there is addingPercentEncoding
let originalString = "test/test"
let escapedString = originalString.addingPercentEncoding(withAllowedCharacters: .urlHostAllowed)
print(escapedString!)
Output:
test%2Ftest
Swift 1
In iOS 7 and above there is stringByAddingPercentEncodingWithAllowedCharacters
var originalString = "test/test"
var escapedString = originalString.stringByAddingPercentEncodingWithAllowedCharacters(.URLHostAllowedCharacterSet())
println("escapedString: \(escapedString)")
Output:
test%2Ftest
The following are useful (inverted) character sets:
URLFragmentAllowedCharacterSet "#%<>[\]^`{|}
URLHostAllowedCharacterSet "#%/<>?#\^`{|}
URLPasswordAllowedCharacterSet "#%/:<>?#[\]^`{|}
URLPathAllowedCharacterSet "#%;<>?[\]^`{|}
URLQueryAllowedCharacterSet "#%<>[\]^`{|}
URLUserAllowedCharacterSet "#%/:<>?#[\]^`
If you want a different set of characters to be escaped create a set:
Example with added "=" character:
var originalString = "test/test=42"
var customAllowedSet = NSCharacterSet(charactersInString:"=\"#%/<>?#\\^`{|}").invertedSet
var escapedString = originalString.stringByAddingPercentEncodingWithAllowedCharacters(customAllowedSet)
println("escapedString: \(escapedString)")
Output:
test%2Ftest%3D42
Example to verify ascii characters not in the set:
func printCharactersInSet(set: NSCharacterSet) {
var characters = ""
let iSet = set.invertedSet
for i: UInt32 in 32..<127 {
let c = Character(UnicodeScalar(i))
if iSet.longCharacterIsMember(i) {
characters = characters + String(c)
}
}
print("characters not in set: \'\(characters)\'")
}
You can use URLComponents to avoid having to manually percent encode your query string:
let scheme = "https"
let host = "www.google.com"
let path = "/search"
let queryItem = URLQueryItem(name: "q", value: "Formula One")
var urlComponents = URLComponents()
urlComponents.scheme = scheme
urlComponents.host = host
urlComponents.path = path
urlComponents.queryItems = [queryItem]
if let url = urlComponents.url {
print(url) // "https://www.google.com/search?q=Formula%20One"
}
extension URLComponents {
init(scheme: String = "https",
host: String = "www.google.com",
path: String = "/search",
queryItems: [URLQueryItem]) {
self.init()
self.scheme = scheme
self.host = host
self.path = path
self.queryItems = queryItems
}
}
let query = "Formula One"
if let url = URLComponents(queryItems: [URLQueryItem(name: "q", value: query)]).url {
print(url) // https://www.google.com/search?q=Formula%20One
}
Swift 4 & 5
To encode a parameter in URL I find using .alphanumerics character set the easiest option:
let urlEncoded = value.addingPercentEncoding(withAllowedCharacters: .alphanumerics)
let url = "http://www.example.com/?name=\(urlEncoded!)"
Using any of the standard Character Sets for URL Encoding (like .urlQueryAllowed or .urlHostAllowed) won't work, because they do not exclude = or & characters.
Note that by using .alphanumerics it will encode some characters that do not need to be encoded (like -, ., _ or ~ -– see 2.3. Unreserved characters in RFC 3986). I find using .alphanumerics simpler than constructing a custom character set and do not mind some additional characters to be encoded. If that bothers you, construct a custom character set as is described in How to percent encode a URL String, like for example:
// Store allowed character set for reuse (computed lazily).
private let urlAllowed: CharacterSet =
.alphanumerics.union(.init(charactersIn: "-._~")) // as per RFC 3986
extension String {
var urlEncoded: String? {
return addingPercentEncoding(withAllowedCharacters: urlAllowed)
}
}
let url = "http://www.example.com/?name=\(value.urlEncoded!)"
Warning: The urlEncoded parameter is force unwrapped. For invalid unicode string it might crash. See Why is the return value of String.addingPercentEncoding() optional?. Instead of force unwrapping urlEncoded! you can use urlEncoded ?? "" or if let urlEncoded = urlEncoded { ... }.
Swift 5:
extension String {
var urlEncoded: String? {
let allowedCharacterSet = CharacterSet.alphanumerics.union(CharacterSet(charactersIn: "~-_."))
return self.addingPercentEncoding(withAllowedCharacters: allowedCharacterSet)
}
}
print("\u{48}ello\u{9}world\u{7}\u{0}".urlEncoded!) // prints Hello%09world%07%00
print("The string ü#foo-bar".urlEncoded!) // prints The%20string%20%C3%BC%40foo-bar
Swift 3:
let originalString = "http://www.ihtc.cc?name=htc&title=iOS开发工程师"
1. encodingQuery:
let escapedString = originalString.addingPercentEncoding(withAllowedCharacters:NSCharacterSet.urlQueryAllowed)
result:
"http://www.ihtc.cc?name=htc&title=iOS%E5%BC%80%E5%8F%91%E5%B7%A5%E7%A8%8B%E5%B8%88"
2. encodingURL:
let escapedString = originalString.addingPercentEncoding(withAllowedCharacters: .urlHostAllowed)
result:
"http:%2F%2Fwww.ihtc.cc%3Fname=htc&title=iOS%E5%BC%80%E5%8F%91%E5%B7%A5%E7%A8%8B%E5%B8%88"
Swift 4 & 5 (Thanks #sumizome for suggestion. Thanks #FD_ and #derickito for testing)
var allowedQueryParamAndKey = NSCharacterSet.urlQueryAllowed
allowedQueryParamAndKey.remove(charactersIn: ";/?:#&=+$, ")
paramOrKey.addingPercentEncoding(withAllowedCharacters: allowedQueryParamAndKey)
Swift 3
let allowedQueryParamAndKey = NSCharacterSet.urlQueryAllowed.remove(charactersIn: ";/?:#&=+$, ")
paramOrKey.addingPercentEncoding(withAllowedCharacters: allowedQueryParamAndKey)
Swift 2.2 (Borrowing from Zaph's and correcting for url query key and parameter values)
var allowedQueryParamAndKey = NSCharacterSet(charactersInString: ";/?:#&=+$, ").invertedSet
paramOrKey.stringByAddingPercentEncodingWithAllowedCharacters(allowedQueryParamAndKey)
Example:
let paramOrKey = "https://some.website.com/path/to/page.srf?a=1&b=2#top"
paramOrKey.addingPercentEncoding(withAllowedCharacters: allowedQueryParamAndKey)
// produces:
"https%3A%2F%2Fsome.website.com%2Fpath%2Fto%2Fpage.srf%3Fa%3D1%26b%3D2%23top"
This is a shorter version of Bryan Chen's answer. I'd guess that urlQueryAllowed is allowing the control characters through which is fine unless they form part of the key or value in your query string at which point they need to be escaped.
Swift 4:
It depends by the encoding rules followed by your server.
Apple offer this class method, but it don't report wich kind of RCF protocol it follows.
var escapedString = originalString.addingPercentEncoding(withAllowedCharacters: .urlHostAllowed)!
Following this useful tool you should guarantee the encoding of these chars for your parameters:
$ (Dollar Sign) becomes %24
& (Ampersand) becomes %26
+ (Plus) becomes %2B
, (Comma) becomes %2C
: (Colon) becomes %3A
; (Semi-Colon) becomes %3B
= (Equals) becomes %3D
? (Question Mark) becomes %3F
# (Commercial A / At) becomes %40
In other words, speaking about URL encoding, you should following the RFC 1738 protocol.
And Swift don't cover the encoding of the + char for example, but it works well with these three # : ? chars.
So, to correctly encoding each your parameter , the .urlHostAllowed option is not enough, you should add also the special chars as for example:
encodedParameter = parameter.replacingOccurrences(of: "+", with: "%2B")
Hope this helps someone who become crazy to search these informations.
Everything is same
var str = CFURLCreateStringByAddingPercentEscapes(
nil,
"test/test",
nil,
"!*'();:#&=+$,/?%#[]",
CFStringBuiltInEncodings.UTF8.rawValue
)
// test%2Ftest
Swift 4.2
A quick one line solution. Replace originalString with the String you want to encode.
var encodedString = originalString.addingPercentEncoding(withAllowedCharacters: CharacterSet(charactersIn: "!*'();:#&=+$,/?%#[]{} ").inverted)
Online Playground Demo
This is working for me in Swift 5. The usage case is taking a URL from the clipboard or similar which may already have escaped characters but which also contains Unicode characters which could cause URLComponents or URL(string:) to fail.
First, create a character set that includes all URL-legal characters:
extension CharacterSet {
/// Characters valid in at least one part of a URL.
///
/// These characters are not allowed in ALL parts of a URL; each part has different requirements. This set is useful for checking for Unicode characters that need to be percent encoded before performing a validity check on individual URL components.
static var urlAllowedCharacters: CharacterSet {
// Start by including hash, which isn't in any set
var characters = CharacterSet(charactersIn: "#")
// All URL-legal characters
characters.formUnion(.urlUserAllowed)
characters.formUnion(.urlPasswordAllowed)
characters.formUnion(.urlHostAllowed)
characters.formUnion(.urlPathAllowed)
characters.formUnion(.urlQueryAllowed)
characters.formUnion(.urlFragmentAllowed)
return characters
}
}
Next, extend String with a method to encode URLs:
extension String {
/// Converts a string to a percent-encoded URL, including Unicode characters.
///
/// - Returns: An encoded URL if all steps succeed, otherwise nil.
func encodedUrl() -> URL? {
// Remove preexisting encoding,
guard let decodedString = self.removingPercentEncoding,
// encode any Unicode characters so URLComponents doesn't choke,
let unicodeEncodedString = decodedString.addingPercentEncoding(withAllowedCharacters: .urlAllowedCharacters),
// break into components to use proper encoding for each part,
let components = URLComponents(string: unicodeEncodedString),
// and reencode, to revert decoding while encoding missed characters.
let percentEncodedUrl = components.url else {
// Encoding failed
return nil
}
return percentEncodedUrl
}
}
Which can be tested like:
let urlText = "https://www.example.com/폴더/search?q=123&foo=bar&multi=eggs+and+ham&hangul=한글&spaced=lovely%20spam&illegal=<>#top"
let url = encodedUrl(from: urlText)
Value of url at the end: https://www.example.com/%ED%8F%B4%EB%8D%94/search?q=123&foo=bar&multi=eggs+and+ham&hangul=%ED%95%9C%EA%B8%80&spaced=lovely%20spam&illegal=%3C%3E#top
Note that both %20 and + spacing are preserved, Unicode characters are encoded, the %20 in the original urlText is not double encoded, and the anchor (fragment, or #) remains.
Edit: Now checking for validity of each component.
For Swift 5 to endcode string
func escape(string: String) -> String {
let allowedCharacters = string.addingPercentEncoding(withAllowedCharacters: CharacterSet(charactersIn: ":=\"#%/<>?#\\^`{|}").inverted) ?? ""
return allowedCharacters
}
How to use ?
let strEncoded = self.escape(string: "http://www.edamam.com/ontologies/edamam.owl#recipe_e2a1b9bf2d996cbd9875b80612ed9aa4")
print("escapedString: \(strEncoded)")
Had need of this myself, so I wrote a String extension that both allows for URLEncoding strings, as well as the more common end goal, converting a parameter dictionary into "GET" style URL Parameters:
extension String {
func URLEncodedString() -> String? {
var escapedString = self.addingPercentEncoding(withAllowedCharacters: .urlHostAllowed)
return escapedString
}
static func queryStringFromParameters(parameters: Dictionary<String,String>) -> String? {
if (parameters.count == 0)
{
return nil
}
var queryString : String? = nil
for (key, value) in parameters {
if let encodedKey = key.URLEncodedString() {
if let encodedValue = value.URLEncodedString() {
if queryString == nil
{
queryString = "?"
}
else
{
queryString! += "&"
}
queryString! += encodedKey + "=" + encodedValue
}
}
}
return queryString
}
}
Enjoy!
This one is working for me.
func stringByAddingPercentEncodingForFormData(plusForSpace: Bool=false) -> String? {
let unreserved = "*-._"
let allowed = NSMutableCharacterSet.alphanumericCharacterSet()
allowed.addCharactersInString(unreserved)
if plusForSpace {
allowed.addCharactersInString(" ")
}
var encoded = stringByAddingPercentEncodingWithAllowedCharacters(allowed)
if plusForSpace {
encoded = encoded?.stringByReplacingOccurrencesOfString(" ", withString: "+")
}
return encoded
}
I found above function from this link: http://useyourloaf.com/blog/how-to-percent-encode-a-url-string/.
let Url = URL(string: urlString.addingPercentEncoding(withAllowedCharacters: .urlQueryAllowed) ?? "")
None of these answers worked for me. Our app was crashing when a url contained non-English characters.
let unreserved = "-._~/?%$!:"
let allowed = NSMutableCharacterSet.alphanumeric()
allowed.addCharacters(in: unreserved)
let escapedString = urlString.addingPercentEncoding(withAllowedCharacters: allowed as CharacterSet)
Depending on the parameters of what you are trying to do, you may want to just create your own character set. The above allows for english characters, and -._~/?%$!:
What helped me was that I created a separate NSCharacterSet and used it on an UTF-8 encoded string i.e. textToEncode to generate the required result:
var queryCharSet = NSCharacterSet.urlQueryAllowed
queryCharSet.remove(charactersIn: "+&?,:;#+=$*()")
let utfedCharacterSet = String(utf8String: textToEncode.cString(using: .utf8)!)!
let encodedStr = utfedCharacterSet.addingPercentEncoding(withAllowedCharacters: queryCharSet)!
let paramUrl = "https://api.abc.eu/api/search?device=true&query=\(escapedStr)"
SWIFT 4.2
Sometimes this happened just because there is space in slug OR absence of URL encoding for parameters passing through API URL.
let myString = self.slugValue
let csCopy = CharacterSet(bitmapRepresentation: CharacterSet.urlPathAllowed.bitmapRepresentation)
let escapedString = myString!.addingPercentEncoding(withAllowedCharacters: csCopy)!
//always "info:hello%20world"
print(escapedString)
NOTE : Don't forget to explore about bitmapRepresentation.
version:Swift 5
// space convert to +
let mstring = string.replacingOccurrences(of: " ", with: "+")
// remove special character
var allowedQueryParamAndKey = NSCharacterSet.urlQueryAllowed
allowedQueryParamAndKey.remove(charactersIn: "!*'\"();:#&=+$,/?%#[]%")
return mstring.addingPercentEncoding(withAllowedCharacters: allowedQueryParamAndKey) ?? mstring
Swift 5
You can try .afURLQueryAllowed option if you want to encode string like below
let testString = "6hAD9/RjY+SnGm&B"
let escodedString = testString.addingPercentEncoding(withAllowedCharacters: .afURLQueryAllowed)
print(escodedString!)
//encoded string will be like en6hAD9%2FRjY%2BSnGm%26B
Is there a way to convert a hexadecimal string to base64 in Swift? For example, I would like to convert:
BA5E64C0DE
to:
ul5kwN4=
It's possible to convert a normal string to base64 by using:
let hex: String = "BA5E64C0DE"
let utf8str: NSData = hex.dataUsingEncoding(NSUTF8StringEncoding)!
let base64Encoded: NSString = utf8str.base64EncodedStringWithOptions(NSDataBase64EncodingOptions(rawValue: 0))
let base64: String = (base64Encoded as String)
But this would give a result of:
QkE1RTY0QzBERQ==
Because it's just treating the hex as a normal UTF-8 String, and not hexadecimal.
It's possible to convert it to base64 correctly by looping through every six hex characters and converting it to it's four respective base64 characters, but this would be highly inefficient, and is just plain stupid (there would need to be 17,830,160 if statements):
if(hex == "000000"){base64+="AAAA"}
else if(hex == "000001"){base64+="AAAB"}
else if(hex == "000002"){base64+="AAAC"}
else if(hex == "BA5E64"){base64+="ul5k"}
//...
It would be nice if there was something like this:
let hex: String = "BA5E64C0DE"
let data: NSData = hex.dataUsingEncoding(NSHexadecimalEncoding)!
let base64Encoded: NSString = data.base64EncodedStringWithOptions(NSDataBase64EncodingOptions(rawValue: 0))
let base64: String = (base64Encoded as String)
But sadly, there's no NSHexadecimalEncoding. Is there any efficient way to convert a hexadecimal string to it's base64 representation in Swift?
The base-64 string, "ul5kwN4=", translates to a binary NSData consisting of of five bytes BA, 5E, 64, C0, and DE.
Now, if you really had a string with the hexadecimal representation, you could convert it to a binary NSData using a routine like the one here: https://stackoverflow.com/a/26502285/1271826
Once you have a NSData, you could build your base 64 string:
let hexString = "BA5E64C0DE"
let binaryData = hexString.dataFromHexadecimalString()
let base64String = binaryData?.base64EncodedStringWithOptions(nil)
That generates the desired output, ul5kwN4=.
First, convert Hex String to Data using the following routine. (Worked in Swift 3.0.2)
extension String {
/// Expanded encoding
///
/// - bytesHexLiteral: Hex string of bytes
/// - base64: Base64 string
enum ExpandedEncoding {
/// Hex string of bytes
case bytesHexLiteral
/// Base64 string
case base64
}
/// Convert to `Data` with expanded encoding
///
/// - Parameter encoding: Expanded encoding
/// - Returns: data
func data(using encoding: ExpandedEncoding) -> Data? {
switch encoding {
case .bytesHexLiteral:
guard self.characters.count % 2 == 0 else { return nil }
var data = Data()
var byteLiteral = ""
for (index, character) in self.characters.enumerated() {
if index % 2 == 0 {
byteLiteral = String(character)
} else {
byteLiteral.append(character)
guard let byte = UInt8(byteLiteral, radix: 16) else { return nil }
data.append(byte)
}
}
return data
case .base64:
return Data(base64Encoded: self)
}
}
}
Then, convert Data to Base64 String using Data.base64EncodedString(options:).
Usage
let base64 = "BA5E64C0DE".data(using: .bytesHexLiteral)?.base64EncodedString()
if let base64 = base64 {
print(base64)
// Prints "ul5kwN4="
}