Trying to pass a string to URLComponents's percentEncodedPath crashes the app if the string is not a valid percent encoded path.
Is there a way to tell if the string is a valid percent encoded path before I set it? I am having trouble figuring this one out.
var urlComponents = URLComponents()
urlComponents.percentEncodedPath = "/some failing path/" // <- This crashes
You can do something like this:
extension String {
var isPercentEncoded: Bool {
let decoded = self.removingPercentEncoding
return decoded != nil && decoded != self
}
}
The rationale:
If the String is URL-encoded, removingPercentEncoding will decode it, and hence decoded will be different from self
If the String contains a percent (but is not URL-Encoded), removingPercentEncoding will fail, and hence decoded will be nil
Otherwise string will remain unmodified
Example:
let failingPath = "/some failing path/"
let succeedingPath = "%2Fsome+succeeding+path%2F"
let doubleEncoded = "%252Fsome%2Bfailing%2Bpath%252F"
let withPercent = "a%b"
failingPath.isPercentEncoded // returns false
succeedingPath.isPercentEncoded // returns true
doubleEncoded.isPercentEncoded // returns true
withPercent.isPercentEncoded // returns false
One disadvantage is that you are on a mercy of removingPercentEncoding and also this is a "mechanical" interpretation of the string, which may not take into account the implementation intent. For example a%20b will be interpreted as URL-encoded. But what if your app expects that string as a decoded one (and so it needs to be encoded further)?
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'm trying to get an understanding of some code I came across recently.
In an answer to a question here https://stackoverflow.com/a/51173170/1162328, the author made use of a String with a format specifier when looping over files in the documentDirectory. Can anyone shed some light on what %#/%# is actually doing?
for fileName in fileNames {
let tempPath = String(format: "%#/%#", path, fileName)
// Check for specific file which you don't want to delete. For me .sqlite files
if !tempPath.contains(".sql") {
try fileManager.removeItem(atPath: tempPath)
}
}
Reading the Apple documentation archive for Formatting Basics I came across this:
In format strings, a ‘%’ character announces a placeholder for a value, with the characters that follow determining the kind of value expected and how to format it. For example, a format string of "%d houses" expects an integer value to be substituted for the format expression '%d'. NSString supports the format characters defined for the ANSI C functionprintf(), plus ‘#’ for any object.
What exactly then, is %#/%# doing?
Each format specifier is replaced by one of the following arguments (usually in the same order, although that can be controlled with positional arguments). So in your case, the first %# is replaced by path and the second %# is replaced by fileName. Example:
let path = "/path/to/dir"
let fileName = "foo.txt"
let tempPath = String(format: "%#/%#", path, fileName)
print(tempPath) // /path/to/dir/foo.txt
The preferred way to build file names and paths is to use the corresponding URL methods instead of string manipulation. Example:
let pathURL = URL(fileURLWithPath: path)
let tempURL = pathURL.appendingPathComponent(fileName)
if tempURL.pathExtension != "sql" {
try FileManager.default.removeItem(at: tempURL)
}
%# is something similar to %d or anything like that. This is the way of string interpolation in Swift.
To be exact %# is placeholder for object - used in Objective-C A LOT. Since NSString * was object (now it is only String), it was used to insert NSString * into another NSString *.
Also given code is just rewritten objective-c code which was something like
NSString *tempPath = [NSString stringWithFormat:#"%#/%#", path, filename];
which can be rewritten in swift:
let tempPath = path + "/" + fileName
Also, given path = "Test" and fileName = "great" will give output Test/great.
One more note: %# is as good as dangerous. You can put UITableView as well as String in it. It will use description property for inserting into string.
I am stuck at getting a string from html body
<html><head>
<title>Uaeexchange Mobile Application</title></head><body>
<div id='ourMessage'>
49.40:51.41:50.41
</div></body></html>
I Would like to get the string containing 49.40:51.41:50.41 . I don't want to do it by string advance or index. Can I get this string by specifying I need only numbers,dot(.) and colon(:) in swift. I mean some numbers and some special characters?
I tried
let stringArray = response.componentsSeparatedByCharactersInSet(
NSCharacterSet.decimalDigitCharacterSet().invertedSet)
let newString = stringArray.joinWithSeparator("")
print("Trimmed\(newString)and count\(newString.characters.count)")
but this obviously trims away dot and colon too. any suggestions friends?
The simple answer to your question is that you need to include "." & ":" in the set that you want to keep.
let response: String = "<html><head><title>Uaeexchange Mobile Application</title></head><body><div id='ourMessage'>49.40:51.41:50.41</div></body></html>"
var s: CharacterSet = CharacterSet.decimalDigits
s.insert(charactersIn: ".:")
let stringArray: [String] = response.components(separatedBy: s.inverted)
let newString: String = stringArray.joined(separator: "")
print("Trimmed '\(newString)' and count=\(newString.characters.count)")
// "Trimmed '49.40:51.41:50.41' and count=17\n"
Without more information on what else your response might be, I can't really give a better answer, but fundamentally this is not a good solution. What if the response had been
<html><head><title>Uaeexchange Mobile Application</title></head><body>
<div id='2'>Some other stuff: like this</div>
<div id='ourMessage'>49.40:51.41:50.41</div>
</body></html>
Using a replace/remove solution to this is a hack, not an algorithm - it will work until it doesn't.
I think you should probably be looking for the <div id='ourMessage'> and reading from there to the next <, but again, we'd need more information on the specification of the format of the response.
I'd recommend to use an HTML parser, nevertheless this is a simple solution with regular expression:
let extractedString = response.replacingOccurrences(of: "[^\\d:.]+", with: "", options: .regularExpression)
Or the positive regex search which is more code but also more reliable:
let pattern = ">\\s?([\\d:.]+)\\s?<"
let regex = try! NSRegularExpression(pattern: pattern)
if let match = regex.firstMatch(in: response, range: NSMakeRange(0, response.utf8.count)) {
let range = match.rangeAt(1)
let startIndex = response.index(response.startIndex, offsetBy: range.location)
let endIndex = response.index(startIndex, offsetBy: range.length)
let extractedString = response.substring(with: startIndex..<endIndex)
print(extractedString)
}
While the simple (negative) regex search removes all characters which don't match digits, dots and colons the positive search considers also the closing (>) and opening tags (<) around the desired result so an accidental digit, dot or colon doesn't match the pattern.
You can also use the String.replacingOccurrences() method in other ways, without regex, as follows:
import Foundation
var response: String = "<html><head><title>Uaeexchange Mobile Application</title></head><body><div id='ourMessage'>49.40:51.41:50.41</div></body></html>"
let charsNotToBeTrimmed = (0...9).map{String($0)} + ["." ,":"] // you can add any character you want here, that's the advantage
for i in response.characters{
if !charsNotToBeTrimmed.contains(String(i)){
response = response.replacingOccurrences(of: String(i), with: "")
}
}
print(response)
Basically, this creates an array of characters which should not be trimmed and if a character is not out there, it gets removed in the for-loop
But you have to be warned that what you're trying to do isn't quite right...