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.
Related
Need to remove part of the string based on brackets.
For ex:
let str = "My Account (_1234)"
I wanted to remove inside brackets (_1234) and the result should be string My Account
Expected Output:
My Account
How can we achieve this with regex format or else using swift default class,need support on this.
Tried with separatedBy but it splits string into array but that is not the expected output.
str.components(separatedBy: "(")
Use replacingOccurrences(…)
let result = str.replacingOccurrences(of: #"\(.*\)"#,
with: "",
options: .regularExpression)
.trimmingCharacters(in: .whitespaces))
Regex ist not necessary. Just get the range of ( and extract the substring up to the lower bound of the range
let str = "My Account (_1234)"
if let range = str.range(of: " (") {
let account = String(str[..<range.lowerBound])
print(account)
}
I'm parsing an XML doc (using XMLParser) and some of the values have php-like placeholders, e.g. %1$s, and I would like to convert those to {x-1}.
Examples:
%1$s ---> {0}
%2$s ---> {1}
I'm doing this in a seemingly hacky way, using regex:
But there must be a better implementation of this regex.
Consider a string:
let str = "lala fawesfgeksgjesk 3rf3f %1$s rk32mrk3mfa %2$s fafafczcxz %3$s czcz $#$##%## %4$s qqq %5$s"
Now we're going to extract the integer strings between strings % and $s:
let regex = try! NSRegularExpression(pattern: "(?<=%)[^$s]+")
let range = NSRange(location: 0, length: str.utf16.count)
let matches = regex.matches(in: str, options: [], range: range)
matches.map {
print(String(str[Range($0.range, in: str)!]))
}
Works quite fine. The issue is that the "4" value got mixed up because of the preceding random strings before the %4$s.
Prints:
1
2
3
## %4
5
Is there any better way to do this?
This might not be a very efficient (or swifty :)) way but it gets the job done. What it does is that it searches for a given reg ex and uses the matched substring to extract the numeric value and decrease it and then perform a simple replace between the substring and a newly constructed placeholder value. This is executed in a loop until no more matches are found.
let pattern = #"%(\d*)\$s"#
while let range = str.range(of: pattern, options: .regularExpression) {
let placeholder = str[range]
let number = placeholder.trimmingCharacters(in: CharacterSet(charactersIn: "0123456789.").inverted)
if let value = Int(number) {
str = str.replacingOccurrences(of: placeholder, with: "{\(value - 1)}")
}
}
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: "Hey #username that's funny". For a given string, how can I search the string to return all ranges of string with first character # and last character to get the username?
I suppose I can get all indexes of # and for each, get the substringToIndex of the next space character, but wondering if there's an easier way.
If your username can contain only letters and numbers, you can use regular expression for that:
let s = "Hey #username123 that's funny"
if let r = s.rangeOfString("#\\w+", options: NSStringCompareOptions.RegularExpressionSearch) {
let name = s.substringWithRange(r) // #username123"
}
#Vladimir's answer is correct, but if you're trying to find multiple occurrences of "username", this should also work:
let s = "Hey #username123 that's funny"
let ranges: [NSRange]
do {
// Create the regular expression.
let regex = try NSRegularExpression(pattern: "#\\w+", options: [])
// Use the regular expression to get an array of NSTextCheckingResult.
// Use map to extract the range from each result.
ranges = regex.matchesInString(s, options: [], range: NSMakeRange(0, s.characters.count)).map {$0.range}
}
catch {
// There was a problem creating the regular expression
ranges = []
}
for range in ranges {
print((s as NSString).substringWithRange(range))
}
Suppose I have a string "10.9.1.1", I want to get substring "10.9". How can I achieve this?
So far I have the following:
var str = "10.9.1.1"
let range = str.rangeOfString(".",options: .RegularExpressionSearch)!
let rangeOfDecimal = Range(start:str.startIndex,end:range.endIndex)
var subStr = str.subStringWithRange(rangeOfDecimal)
But this will only return 10.
Actually your code returns "1" only, because "." in a regular
expression pattern matches any character.
The correct pattern would be
\d+ one ore more digits
\. a literal dot
\d+ one or more digits
In a Swift string, you have to escape the backslashes as "\\":
let str = "10.9.1.1"
if let range = str.rangeOfString("\\d+\\.\\d+",options: .RegularExpressionSearch) {
let subStr = str.substringWithRange(range)
println(subStr) // "10.9"
}