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 receive value from API response like below
{
"first_name" = "active EQ 'true'";
}
I need to send that value as a query parameter in another API. But when sending the request I see that URL is not being encoded. I am expecting that active EQ 'true' should be URL encoded as active%20EQ%20%27true%27.
I tried addingPercentEncoding(withAllowedCharacters: .urlQueryAllowed) and it resulted in "active%20EQ%20\'true\'. I see backslash is being added as escape character but not encoding as expected.
let k = "active EQ 'true'"
let j = k.addingPercentEncoding(withAllowedCharacters: .urlQueryAllowed)
print("j is \(j)")
I expected the output as active%20EQ%20%27true%27, but the actual output is active%20EQ%20\'true\'
Omit " ' " from your string first. Check following code :
let k = "active EQ 'true' "
let i = k.replacingOccurrences(of: "'", with: "")
let j = i.addingPercentEncoding(withAllowedCharacters: .urlQueryAllowed)
print("j is \(j!)")
Thanks for all suggestions and by EOD I don't want to encode single quotes as the API is working fine with that and value to API should be sent with single quotes only. I was just trying to know the possibilities of encoding this type of strings.
Try this, I checked it and its working fine.
let k = "active EQ 'true'"
let j = escape(k)
print("j is \(j)")
func escape(_ string: String) -> String {
let generalDelimitersToEncode = ":#[]#" // does not include "?" or "/" due to RFC 3986 - Section 3.4
let subDelimitersToEncode = "!$&'()*+,;="
var allowedCharacterSet = CharacterSet.urlQueryAllowed
allowedCharacterSet.remove(charactersIn: "\(generalDelimitersToEncode)\(subDelimitersToEncode)")
return string.addingPercentEncoding(withAllowedCharacters: allowedCharacterSet) ?? string
}
Result: j is active%20EQ%20%27true%27
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)"
I have a string like "http://example.com/a/b/c". I want to transform the string to "httpexamplecomabc" in order to save it as the file name. I have tried
let result = str.stringByTrimmingCharactersInSet(NSCharacterSet(charactersInString: "//"))
and
let result = str.stringByTrimmingCharactersInSet(NSCharacterSet(charactersInString: "/"))
but neither works. Any Idea on how to remove "/"? Thanks
You can replace occurrence of "/ " to "" by using
let mm = "http://example.com/a/b/c"
let newString = mm.stringByReplacingOccurrencesOfString("/", withString: "")
print(newString) // http:example.comabc