URLComponent percentEncodedPath - Is value valid before setting it - swift

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)?

Related

Xcode Swift URL with "|" character [duplicate]

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

How can I convert unicode values in a string "U+XXXXXX" to the corresponding characters in Swift?

I accept from backend the following json
{
"id": "f33919f6-3554-4246-9e78-bca3a690c119",
"title": "Category3",
"slug": "category3",
"hex_up": "#eb4034",
"hex_down": "#80302a",
"emoji": "U+1F602",
"parent_id": "aa3f651b-f068-4ae1-a9d8-a18a9945b111"
}
There is a field "emoji": "U+1F602",
I need show emoji icon like 😃 in UILabel
I tried to google and found results like
let scalarValue = UnicodeScalar(emojiString)
let myString = String(scalarValue!)
Unfortunately app crashes at the second line.
Thanks for your answers.
There's no U+... syntax in Swift. (There is a \u{...} syntax that does the same thing, but it's not necessary here.)
You'll need to parse the String yourself:
func parseUnicode(_ string: String) -> String? {
guard string.hasPrefix("U+"), // Make sure it's a U+ string
let value = Int(string.dropFirst(2), radix: 16), // Convert to Int
let scalar = UnicodeScalar(value) // Convert to UnicodeScalar
else { return nil }
return String(scalar) // Convert to String
}
if let myString = parseUnicode(emoji) { ... }
Don't use ! here. The U+... string may be invalid, and you wouldn't want to crash in that case.
You can simply apply a string transform from "Hex/Unicode" to "Any" (a set of all characters):
"U+1F602".applyingTransform(.init("Hex/Unicode-Any"), reverse: false) // "😂"
or as instance properties of StringProtocol to encode/decode from/to hexa unicode:
extension StringTransform {
static let unicodeToAny: Self = .init("Hex/Unicode-Any")
static let anyToUnicode: Self = .init("Any-Hex/Unicode")
}
extension StringProtocol {
var decodingHexaUnicode: String {
applyingTransform(.unicodeToAny, reverse: false)!
}
var encodingHexaUnicode: String {
applyingTransform(.anyToUnicode, reverse: false)!
}
}
Usage:
let hexaUnicode = "U+1F602"
let emoji = hexaUnicode.decodingHexaUnicode // "😂"
let unicodeFromEmoji = emoji.encodingHexaUnicode // "U+1F602"
The reason your app crashed is due to the fact that the scalarValue you attempted to initialize is nil, and you're force-unwrapping using (!) that nil value on line 2. Rob's answer shows how to unwrap the optional safely.
You can get the emoji by using the value following the U+. So you'll need to drop the first two characters of the string. So use this code to accomplish that:
let parsedEmoji = emojiString.substring(from:2)
Now you'll convert that emoji unicode using the code below.
let emoji = String(UnicodeScalar(Int(parsedEmojiHex,radix: 16)!)!)
print(emoji)

Using String addingPercentEncoding to encode a string as query param

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"

Conversion stringByAddingPercentEncodingWithAllowedCharacters(.URLHostAllowedCharacterSet()) from Swift2 to Swift3 [duplicate]

I was using this, in Swift 1.2
let urlwithPercentEscapes = myurlstring.stringByAddingPercentEscapesUsingEncoding(NSUTF8StringEncoding)
This now gives me a warning asking me to use
stringByAddingPercentEncodingWithAllowedCharacters
I need to use a NSCharacterSet as an argument, but there are so many and I cannot determine what one will give me the same outcome as the previously used method.
An example URL I want to use will be like this
http://www.mapquestapi.com/geocoding/v1/batch?key=YOUR_KEY_HERE&callback=renderBatch&location=Pottsville,PA&location=Red Lion&location=19036&location=1090 N Charlotte St, Lancaster, PA
The URL Character Set for encoding seems to contain sets the trim my
URL. i.e,
The path component of a URL is the component immediately following the
host component (if present). It ends wherever the query or fragment
component begins. For example, in the URL
http://www.example.com/index.php?key1=value1, the path component is
/index.php.
However I don't want to trim any aspect of it.
When I used my String, for example myurlstring it would fail.
But when used the following, then there were no issues. It encoded the string with some magic and I could get my URL data.
let urlwithPercentEscapes = myurlstring.stringByAddingPercentEscapesUsingEncoding(NSUTF8StringEncoding)
As it
Returns a representation of the String using a given encoding to
determine the percent escapes necessary to convert the String into a
legal URL string
Thanks
For the given URL string the equivalent to
let urlwithPercentEscapes = myurlstring.stringByAddingPercentEscapesUsingEncoding(NSUTF8StringEncoding)
is the character set URLQueryAllowedCharacterSet
let urlwithPercentEscapes = myurlstring.stringByAddingPercentEncodingWithAllowedCharacters( NSCharacterSet.URLQueryAllowedCharacterSet())
Swift 3:
let urlwithPercentEscapes = myurlstring.addingPercentEncoding( withAllowedCharacters: .urlQueryAllowed)
It encodes everything after the question mark in the URL string.
Since the method stringByAddingPercentEncodingWithAllowedCharacters can return nil, use optional bindings as suggested in the answer of Leo Dabus.
It will depend on your url. If your url is a path you can use the character set
urlPathAllowed
let myFileString = "My File.txt"
if let urlwithPercentEscapes = myFileString.addingPercentEncoding(withAllowedCharacters: .urlPathAllowed) {
print(urlwithPercentEscapes) // "My%20File.txt"
}
Creating a Character Set for URL Encoding
urlFragmentAllowed
urlHostAllowed
urlPasswordAllowed
urlQueryAllowed
urlUserAllowed
You can create also your own url character set:
let myUrlString = "http://www.mapquestapi.com/geocoding/v1/batch?key=YOUR_KEY_HERE&callback=renderBatch&location=Pottsville,PA&location=Red Lion&location=19036&location=1090 N Charlotte St, Lancaster, PA"
let urlSet = CharacterSet.urlFragmentAllowed
.union(.urlHostAllowed)
.union(.urlPasswordAllowed)
.union(.urlQueryAllowed)
.union(.urlUserAllowed)
extension CharacterSet {
static let urlAllowed = CharacterSet.urlFragmentAllowed
.union(.urlHostAllowed)
.union(.urlPasswordAllowed)
.union(.urlQueryAllowed)
.union(.urlUserAllowed)
}
if let urlwithPercentEscapes = myUrlString.addingPercentEncoding(withAllowedCharacters: .urlAllowed) {
print(urlwithPercentEscapes) // "http://www.mapquestapi.com/geocoding/v1/batch?key=YOUR_KEY_HERE&callback=renderBatch&location=Pottsville,PA&location=Red%20Lion&location=19036&location=1090%20N%20Charlotte%20St,%20Lancaster,%20PA"
}
Another option is to use URLComponents to properly create your url
Swift 3.0 (From grokswift)
Creating URLs from strings is a minefield for bugs. Just miss a single / or accidentally URL encode the ? in a query and your API call will fail and your app won’t have any data to display (or even crash if you didn’t anticipate that possibility). Since iOS 8 there’s a better way to build URLs using NSURLComponents and NSURLQueryItems.
func createURLWithComponents() -> URL? {
var urlComponents = URLComponents()
urlComponents.scheme = "http"
urlComponents.host = "www.mapquestapi.com"
urlComponents.path = "/geocoding/v1/batch"
let key = URLQueryItem(name: "key", value: "YOUR_KEY_HERE")
let callback = URLQueryItem(name: "callback", value: "renderBatch")
let locationA = URLQueryItem(name: "location", value: "Pottsville,PA")
let locationB = URLQueryItem(name: "location", value: "Red Lion")
let locationC = URLQueryItem(name: "location", value: "19036")
let locationD = URLQueryItem(name: "location", value: "1090 N Charlotte St, Lancaster, PA")
urlComponents.queryItems = [key, callback, locationA, locationB, locationC, locationD]
return urlComponents.url
}
Below is the code to access url using guard statement.
guard let url = createURLWithComponents() else {
print("invalid URL")
return nil
}
print(url)
Output:
http://www.mapquestapi.com/geocoding/v1/batch?key=YOUR_KEY_HERE&callback=renderBatch&location=Pottsville,PA&location=Red%20Lion&location=19036&location=1090%20N%20Charlotte%20St,%20Lancaster,%20PA
In Swift 3.1, I am using something like the following:
let query = "param1=value1&param2=" + valueToEncode.addingPercentEncoding(withAllowedCharacters: .alphanumeric)
It's safer than .urlQueryAllowed and the others, because it this will encode every characters other than A-Z, a-z and 0-9. This works better when the value you are encoding may use special characters like ?, &, =, + and spaces.
In my case where the last component was non latin characters I did the following in Swift 2.2:
extension String {
func encodeUTF8() -> String? {
//If I can create an NSURL out of the string nothing is wrong with it
if let _ = NSURL(string: self) {
return self
}
//Get the last component from the string this will return subSequence
let optionalLastComponent = self.characters.split { $0 == "/" }.last
if let lastComponent = optionalLastComponent {
//Get the string from the sub sequence by mapping the characters to [String] then reduce the array to String
let lastComponentAsString = lastComponent.map { String($0) }.reduce("", combine: +)
//Get the range of the last component
if let rangeOfLastComponent = self.rangeOfString(lastComponentAsString) {
//Get the string without its last component
let stringWithoutLastComponent = self.substringToIndex(rangeOfLastComponent.startIndex)
//Encode the last component
if let lastComponentEncoded = lastComponentAsString.stringByAddingPercentEncodingWithAllowedCharacters(NSCharacterSet.alphanumericCharacterSet()) {
//Finally append the original string (without its last component) to the encoded part (encoded last component)
let encodedString = stringWithoutLastComponent + lastComponentEncoded
//Return the string (original string/encoded string)
return encodedString
}
}
}
return nil;
}
}
Swift 4.0
let encodedData = myUrlString.addingPercentEncoding(withAllowedCharacters: CharacterSet.urlHostAllowed)

Replace all Special Characters in String with valid URL characters

I cannot figure out how to replace all special characters in a string and convert it to a string I can use in a URL.
What I am using it for:
I am uploading an image, converting it to base64, and then passing it to the Laravel framework, however the base64 string can contain +, /, \, etc. which changes the meaning of the URL.
I can replace the + sign with the following code:
let withoutPlus = image.stringByReplacingCharactersInRange("+", withString: "%2B")
however then I cannot use that as a NSString to try and change the other characters.
Surely there is a way to just target every single special character and convert it something usable in a URL?
You can use stringByAddingPercentEncodingWithAllowedCharacters to escape characters as needed. You pass it an NSCharacterSet containing the characters that are valid for that string (i.e. the ones you don't want replaced). There's a built-in NSCharacterSet for characters allowed in URL query strings that will get you most of the way there, but it includes + and / so you'll need to remove those from the set. You can do that by making a mutable copy of the set and then calling removeCharactersInString:
let allowedCharacters = NSCharacterSet.URLQueryAllowedCharacterSet().mutableCopy() as NSMutableCharacterSet
allowedCharacters.removeCharactersInString("+/=")
Then you can call stringByAddingPercentEncodingWithAllowedCharacters on your string, passing in allowedCharacters:
let encodedImage = image.stringByAddingPercentEncodingWithAllowedCharacters(allowedCharacters)
Note that it will return an optional String (String?) so you'll probably want to use optional binding:
if let encodedImage = image.stringByAddingPercentEncodingWithAllowedCharacters(allowedCharacters) {
/* use encodedImage here */
} else {
/* stringByAddingPercentEncodingWithAllowedCharacters failed for some reason */
}
Example:
let unencodedString = "abcdef/+\\/ghi"
let allowedCharacters = NSCharacterSet.URLQueryAllowedCharacterSet().mutableCopy() as NSMutableCharacterSet
allowedCharacters.removeCharactersInString("+/=")
if let encodedString = unencodedString.stringByAddingPercentEncodingWithAllowedCharacters(allowedCharacters) {
println(encodedString)
}
Prints:
abcdef%2F%2B%5C%2Fghi
Use
let withoutPlus = image.stringByReplacingOccurrencesOfString("+", withString: "%2B")
rather than image.stringByReplacingCharactersInRange. Note that your call as posted doesn't work, as that method is declared as
func stringByReplacingCharactersInRange(range: Range<String.Index>, withString replacement: String) -> String
and you are not supplying the correct parameters.
You might do better to use POST to send a file, rather than encode it into your URL