I simply need to replace:
<p>, <div>
with
\n<p>, \n<div>
in string with one single pattern replacing. Is it possible?
let string = "<p>hello</p> my <div>Doggy</div>"
let newString = string.replacingOccurrences(of: "<p>", with: "\n<p>").replacingOccurrences(of: "<div>", with: "\n<div>")
is there a better solution with regex?
You can do a regular expression search, with a template in the replacement
string:
let string = "<p>hello</p> my <div>Doggy</div>"
let newString = string.replacingOccurrences(of: "<p>|<div>", with: "\n$0", options: .regularExpression)
For each match, the $0 template is replaced by what actually matched the pattern.
I have Strings with the form string \ string example
"some sting with random length\233"
I want to deletes the last \ and get the value after it, so the result will be
"some sting with random length"
"233"
I tried this code but it's not working
let regex = try! NSRegularExpression(pattern: "\\\s*(\\S[^,]*)$")
if let match = regex.firstMatch(in: string, range: string.nsRange) {
let result = string.substring(with: match.rangeAt(1))
}
You did not correctly adapt the pattern from How to get substring after last occurrence of character in string: Swift IOS to your case. Both instances of the comma must be replaced by a backslash,
and that must be "double-escaped":
let regex = try! NSRegularExpression(pattern: "\\\\\\s*(\\S[^\\\\]*)$")
(once be interpreted as a literal backslash in the regex pattern, and
once more in the definition of a Swift string literal).
However, a simpler solution is to find the last occurrence of the
backslash and extract the suffix from that position:
let string = "some sting with random length\\233"
let separator = "\\" // A single(!) backslash
if let r = string.range(of: separator, options: .backwards) {
let prefix = string.substring(to: r.lowerBound)
let suffix = string.substring(from: r.upperBound)
print(prefix) // some sting with random length
print(suffix) // 233
}
Update for Swift 4:
if let r = string.range(of: separator, options: .backwards) {
let prefix = string[..<r.lowerBound]
let suffix = string[r.upperBound...]
print(prefix) // some sting with random length
print(suffix) // 233
}
prefix and suffix are a String.SubSequence, which can be used
in many places instead of a String. If necessary, create a real
string:
let prefix = String(string[..<r.lowerBound])
let suffix = String(string[r.upperBound...])
You could do this with regex, but I think this solution is better:
yourString.components(separatedBy: "\\").last!
It splits the string with \ as the separator and gets the last split.
I am using xcode 8.3.3 and writing a XCUI test.
I have the following:
let address = XCUIApplication().buttons["URL"].value as! String
Looking in the debugger, I can see the value is:
If I set
expectedURL = "\u{e2}auth.int....net" then it returns:
If I set
expectedURL = "auth.int....net" then it returns:
How can I make the test assertion find the two strings to be equal?
Tried the following, but it doesn't replace the "\u{e2}":
let address = value.components(separatedBy: ",").first!.replacingOccurrences(of: "\u{e2}", with: "")
And also (but it doesn't replace the "\u{e2}"):
let range = Range<String.Index>(uncheckedBounds: (lower: address.startIndex, upper: address.endIndex))
let strippedAddress = address.replacingOccurrences(of:"\\u{e2}", with: "", options: .literal, range: range)
For the assertion, I am using XCTAssertEqual(address, expectedURL)
You can fix it by separating by alphanumeric characters and then joining by an empty string, as shown below.
let myString = address.components(separatedBy: CharacterSet.alphanumerics.inverted).joined(separator: "")
myStringis then equal to authintxxxxxxxxxnet (no "." characters), so you should just be able to change your expected URL to match.
Hope that helps!
I receive an NSAttributedString that contains a NSTextAttachment. I want to remove that attachment, and it looks like it is represented as "\u{ef}" in the string. Printing the unicode scalars of such string, it also seems that unicode scalar for the "\u{ef}" is U\0000fffc.
I tried to do this:
noAttachmentsText = text.replacingOccurrences(of: "\u{ef}", with: "")
with no success, so I'm trying by comparing unicode scalars:
var scalars = Array(text.unicodeScalars)
for scalar in scalars {
// compare `scalar` to `U\0000fffc`
}
but I'm not able either to succeed in the comparison.
How could I do this?
But this code works for me from How do I remove "\U0000fffc" from a string in Swift?
let original = "First part \u{ef} Last part"
let originalRange = Range<String.Index>(start: original.startIndex, end: original.endIndex)
let target = original.stringByReplacingOccurrencesOfString("\u{ef}", withString: "", options: NSStringCompareOptions.LiteralSearch, range: originalRange)
print(target)
Output :
"First part ï Last part"
to
First part Last part
U can use similar code for swift 3 just replace unicode using replacingOccurrences option for exapmle :
func stringTocleanup(str: String) -> String {
var result = str
result = result.replacingOccurrences(of: "\"", with: "\"")
.replacingOccurrences(of: "\u{10}", with: "")
return result
}
I have a string which is "Optional("5")". I need to remove the "" surrounding the 5. I have removed the 'Optional' by doing:
text2 = text2.stringByReplacingOccurrencesOfString("Optional(", withString: "", options: NSStringCompareOptions.LiteralSearch, range: nil)
I am having difficulties removing the " characters as they designate the end of a string in the code.
Swift uses backslash to escape double quotes. Here is the list of escaped special characters in Swift:
\0 (null character)
\\ (backslash)
\t (horizontal tab)
\n (line feed)
\r (carriage return)
\" (double quote)
\' (single quote)
This should work:
text2 = text2.replacingOccurrences(of: "\\", with: "", options: NSString.CompareOptions.literal, range: nil)
Swift 3 and Swift 4:
text2 = text2.textureName.replacingOccurrences(of: "\"", with: "", options: NSString.CompareOptions.literal, range:nil)
Latest documents updated to Swift 3.0.1 have:
Null Character (\0)
Backslash (\\)
Horizontal Tab (\t)
Line Feed (\n)
Carriage Return (\r)
Double Quote (\")
Single Quote (\')
Unicode scalar (\u{n}), where n is between one and eight hexadecimal digits
If you need more details you can take a look to the official docs here
Here is the swift 3 updated answer
var editedText = myLabel.text?.replacingOccurrences(of: "\"", with: "")
Null Character (\0)
Backslash (\\)
Horizontal Tab (\t)
Line Feed (\n)
Carriage Return (\r)
Double Quote (\")
Single Quote (\')
Unicode scalar (\u{n})
To remove the optional you only should do this
println("\(text2!)")
cause if you dont use "!" it takes the optional value of text2
And to remove "" from 5 you have to convert it to NSInteger or NSNumber easy peasy. It has "" cause its an string.
Replacing for Removing is not quite logical.
String.filter allows to iterate a string char by char and keep only true assertion.
Swift 4 & 5
var aString = "Optional(\"5\")"
aString = aString.filter { $0 != "\"" }
> Optional(5)
Or to extend
var aString = "Optional(\"5\")"
let filteredChars = "\"\n\t"
aString = aString.filter { filteredChars.range(of: String($0)) == nil }
> Optional(5)
I've eventually got this to work in the playground, having multiple characters I'm trying to remove from a string:
var otherstring = "lat\" : 40.7127837,\n"
var new = otherstring.stringByTrimmingCharactersInSet(NSCharacterSet.init(charactersInString: "la t, \n \" ':"))
count(new) //result = 10
println(new)
//yielding what I'm after just the numeric portion 40.7127837
If you want to remove more characters for example "a", "A", "b", "B", "c", "C" from string you can do it this way:
someString = someString.replacingOccurrences(of: "[abc]", with: "", options: [.regularExpression, .caseInsensitive])
As Martin R says, your string "Optional("5")" looks like you did something wrong.
dasblinkenlight answers you so it is fine, but for future readers, I will try to add alternative code as:
if let realString = yourOriginalString {
text2 = realString
} else {
text2 = ""
}
text2 in your example looks like String and it is maybe already set to "" but it looks like you have an yourOriginalString of type Optional(String) somewhere that it wasn't cast or use correctly.
I hope this can help some reader.
Swift 5 (working). Only 1 line code.
For removing single / multiple characters.
trimmingCharacters(in: CharacterSet)
In action:
var yourString:String = "(\"This Is: Your String\")"
yourString = yourString.trimmingCharacters(in: ["("," ",":","\"",")"])
print(yourString)
Output:
ThisIsYourString
You are entering a Set that contains characters you're required to trim.
Let's say you have a string:
var string = "potatoes + carrots"
And you want to replace the word "potatoes" in that string with "tomatoes"
string = string.replacingOccurrences(of: "potatoes", with: "tomatoes", options: NSString.CompareOptions.literal, range: nil)
If you print your string, it will now be: "tomatoes + carrots"
If you want to remove the word potatoes from the sting altogether, you can use:
string = string.replacingOccurrences(of: "potatoes", with: "", options: NSString.CompareOptions.literal, range: nil)
If you want to use some other characters in your sting, use:
Null Character (\0)
Backslash (\)
Horizontal Tab (\t)
Line Feed (\n)
Carriage Return (\r)
Double Quote (\")
Single Quote (\')
Example:
string = string.replacingOccurrences(of: "potatoes", with: "dog\'s toys", options: NSString.CompareOptions.literal, range: nil)
Output: "dog's toys + carrots"
If you are getting the output Optional(5) when trying to print the value of 5 in an optional Int or String, you should unwrap the value first:
if value != nil
{ print(value)
}
or you can use this:
if let value = text {
print(value)
}
or in simple just 1 line answer:
print(value ?? "")
The last line will check if variable 'value' has any value assigned to it, if not it will print empty string
You've instantiated text2 as an Optional (e.g. var text2: String?).
This is why you receive Optional("5") in your string.
take away the ? and replace with:
var text2: String = ""
If you are getting the output Optional(5) when trying to print the value of 5 in an optional Int or String, you should unwrap the value first:
if let value = text {
print(value)
}
Now you've got the value without the "Optional" string that Swift adds when the value is not unwrapped before.