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
I have an input string "+20" and I am trying to pass that as query parameter in url.
So I am trying to encode the myInputString by doing
let s1 = myInputString.addingPercentEncoding(withAllowedCharacters: .urlQueryAllowed)
But in the debugger, the string s1 still shows as '+20' instead of '%2B20'
Is there something I did wrong?
As already mentioned by matt + is a legal URL character. If you really need to encode it you would need to create your own custom urlQueryAllowed and subtract the plus sign from it:
extension CharacterSet {
static let allowedCharacters = urlQueryAllowed.subtracting(.init(charactersIn: "+"))
}
let myInputString = "+20"
let s1 = myInputString.addingPercentEncoding(withAllowedCharacters: .allowedCharacters) // "%2B20"
I have a url string (fake)
http://fools.sayers.mine.cs/api/analytics/ces_ssn
And I'd like to create a new URL string
http://fools.sayers.mine.cs/api/
The issue I'm facing is the last parameter ces_ssn can sometimes be anything like ces_fw or adv_let so I can't entirely set an endIndex in my code.
Is there a way to create a function that is dynamic in saying just give me the first 32 characters every time, not matter the endIndex
You have a URL. Use URL. Assuming you know you want to drop the last two parts of the path, you can do:
let myURL = URL(string: "http://fools.sayers.mine.cs/api/analytics/ces_ssn")!
let shortURL = myURL.deletingLastPathComponent().deletingLastPathComponent()
print(shortURL)
Output:
http://fools.sayers.mine.cs/api/
This works for Swift 4
var url = "http://fools.sayers.mine.cs/api/analytics/ces_ssn"
let newStr = url.prefix(32)
or probably the perfect way
if let index = url.range(of: "api/")?.upperBound {
let api = url.prefix(upTo: index)
}
You can use prefix()
"http://fools.sayers.mine.cs/api/analytics/ces_ssn".prefix(32)
That sounds prone to mistake if all of a sudden your URL is https instead of http, for example.
Instead I would do:
let url = URL(string: "http://fools.sayers.mine.cs/api/analytics/ces_ssn")!
let components = url.pathComponents
let scheme = url.scheme!
let host = url.host!
let slash = components.removeFirst()
print (components) // ["api", "analytics", "ces_ssn"]
The components you are interested in are then the 0 and 1 component, and you could reconstruct your URL like this:
let newURL = "\(scheme)://\(host)/\(components[0])/\(components[1])"
print (newURL) // "http://fools.sayers.mine.cs/api/analytics"
I am newbie on Swift and I am building API URL string as follows before pass it to URLSession.
I wonder there is a better way of doing it?
let jsonUrlString = Constants.API_URL + "/" + Constants.PATH + "/"
+ String(page)+"/" + Constants.NUMBER_ITEMS_PER_PAGE
The proper way to build a URL is to use URL and URLComponents. Simply appending strings together is error prone and it does not properly escape special characters that you might have in your values.
Here's one possible solution using URL:
if let baseURL = URL(string: Constants.API_URL) {
let jsonURL = baseURL.appendingPathComponent(Constants.PATH)
.appendingPathComponent(String(page))
.appendingPathComponent(Constants.NUMBER_ITEMS_PER_PAGE)
// use jsonURL with your URLSession
}
Another option with URLComponents (this properly ensures special characters are encoded):
if let baseComps = URLComponents(string: Constants.API_URL) {
var components = baseComps
components.path = "/\(Constants.PATH)/\(page)/\(Constants.NUMBER_ITEMS_PER_PAGE)"
if let jsonURL = components.url {
// use jsonURL with your URLSession
}
}
Also there is another way in swift to build string is called interpolation and mostly this one is used by developers.
If you using this you don't have to take your Int or another type value into string, Because is automatic take value into string.
i.e
let myValue = 3
let intToString = "\(myValue)" // "3"
let doubleValue = 4.5
let doubleToString = "\(doubleValue)" // "4.5"
So you URL will be like as below
let jsonUrlString = "\(Constants.API_URL)/\(Constants.PATH)/\(page)/\(Constants.NUMBER_ITEMS_PER_PAGE)"
When trying to remove the suffix from a filename, I'm only left with the suffix, which is exactly not what I want.
What (how many things) am I doing wrong here:
let myTextureAtlas = SKTextureAtlas(named: "demoArt")
let filename = (myTextureAtlas.textureNames.first?.characters.split{$0 == "."}.map(String.init)[1].replacingOccurrences(of: "\'", with: ""))! as String
print(filename)
This prints png which is the most dull part of the whole thing.
If by suffix you mean path extension, there is a method for this:
let filename = "demoArt.png"
let name = (filename as NSString).deletingPathExtension
// name - "demoArt"
Some people here seem to overlook that a filename can have multiple periods in the name and in that case only the last period separates the file extension. So this.is.a.valid.image.filename.jpg and stripping the extension should return this.is.a.valid.image.filename and not this (as two answers here would produce) or anything else in between. The regex answer works correctly but using a regex for that is a bit overkill (probably 10 times slower than using simple string processing). Here's a generic function that works for everyone:
func stripFileExtension ( _ filename: String ) -> String {
var components = filename.components(separatedBy: ".")
guard components.count > 1 else { return filename }
components.removeLast()
return components.joined(separator: ".")
}
print("1: \(stripFileExtension("foo"))")
print("2: \(stripFileExtension("foo.bar"))")
print("3: \(stripFileExtension("foo.bar.foobar"))")
Output:
foo
foo
foo.bar
You can also split the String using componentsSeparatedBy, like this:
let fileName = "demoArt.png"
var components = fileName.components(separatedBy: ".")
if components.count > 1 { // If there is a file extension
components.removeLast()
return components.joined(separator: ".")
} else {
return fileName
}
To clarify:
fileName.components(separatedBy: ".")
will return an array made up of "demoArt" and "png".
In iOS Array start with 0 and you want name of the file without extension, so you have split the string using ., now the name will store in first object and extension in the second one.
Simple Example
let fileName = "demoArt.png"
let name = fileName.characters.split(".").map(String.init).first
If you don't care what the extension is. This is a simple way.
let ss = filename.prefix(upTo: fileName.lastIndex { $0 == "." } ?? fileName.endIndex))
You may want to convert resulting substring to String after this. With String(ss)
#Confused with Swift 4 you can do this:
let fileName = "demoArt.png"
// or on your specific case:
// let fileName = myTextureAtlas.textureNames.first
let name = String(fileName.split(separator: ".").first!)
print(name)
Additionally you should also unwrapp first but I didn't want to complicate the sample code to solve your problem.
Btw, since I've also needed this recently, if you want to remove a specific suffix you know in advance, you can do something like this:
let fileName = "demoArt.png"
let fileNameExtension = ".png"
if fileName.hasSuffix(fileNameExtension) {
let name = fileName.prefix(fileName.count - fileNameExtension.count)
print(name)
}
How about using .dropLast(k) where k is the number of characters you drop from the suffix ?
Otherwise for removing extensions from path properly from filename, I insist you to use URL and .deletingPathExtension().lastPathComponent.
Maybe a bit overhead but at least it's a rock solid Apple API.
You can also use a Regexp to extract all the part before the last dot like that :
let fileName = "test.png"
let pattern = "^(.*)(\\.[a-zA-Z]+)$"
let regexp = try! NSRegularExpression(pattern: pattern, options: [])
let extractedName = regexp.stringByReplacingMatches(in: fileName, options: [], range: NSMakeRange(0, fileName.characters.count), withTemplate: "$1")
print(extractedName) //test
let mp3Files = ["alarm.mp3", "bubbles.mp3", "fanfare.mp3"]
let ringtonsArray = mp3Files.flatMap { $0.components(separatedBy: ".").first }
You can return a new string removing a definite number of characters from the end.
let fileName = "demoArt.png"
fileName.dropLast(4)
This code returns "demoArt"
One liner:
let stringWithSuffixDropped = fileName.split(separator: ".").dropLast().joined(separator: ".")