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

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)

Related

Check existence of localized string in .strings resource file, fall back to default

I want to find the appropriate localized string based on some runtime variable and fall back to a default string:
// localizable.strings
"com.myapp.text1" = "The default text 1";
"com.myapp.text1#SPECIAL" = "The special text";
"com.myapp.text2" = "The default text 2";
// my code
let key1 = "com.myapp.text1"
let key2 = "com.myapp.text2"
let modifier = "#SPECIAL"
print( NSLocalizedString(key1 + modifier
, value: NSLocalizedString(key1, comment: "")
, comment: "") )
// > "The special text 1"
print( NSLocalizedString(key2 + modifier
, value: NSLocalizedString(key2, comment: "") # the default to fall back to
, comment: "") )
// > "The default text 2"
Nice, that's what I want, try a special variant, fall back to the default.
However, if the option NSShowNonLocalizedStrings in the user defaults is set to true, it fails:
For non-localised strings, an upper-case version of the key will be returned, ignoring the default value. Also an error message is printed in the console (documentation).
So it appears that my solution is working against intended way of using NSLocalizedString.
UserDefaults.standard.set(true, forKey: "NSShowNonLocalizedStrings") # could also be passed as launch option or set via terminal
print( NSLocalizedString(key2 + modifier
, value: NSLocalizedString(key2, comment: "")
, comment: "") )
// > ERROR: com.myapp.text2#MODIFIER not found [...]
// > "COM.MYAPP.TEXT2"
I could work around this by testing for the uppercased version etc. but this would just be a hack that masks the actual issue.
What I would probably need is a test if (bundle.hasLocalizationFor(key2 + modifier)... but to implement such a method I would have to re-implement processing of the strings files including parsing and caching. And that feels wrong in itself.
Question:
Is there some method I am missing to achieve what I am looking for?
Or is the whole idea of special/fallback localization just wrong for the platform?
Thanks to comments by Martin R, I was able to get a reasonably working solution:
static let cache = NSCache<NSString, NSDictionary>()
private func strings(for tableName: String) -> [String: String] {
guard let strings = cache.object(forKey: tableName as NSString) else {
guard let path = Bundle.main.path(forResource: tableName, ofType: "strings")
, let dict = NSDictionary(contentsOfFile: path) as? [String: String] else {
return [:]
}
cache.setObject(dict as NSDictionary, forKey: tableName as NSString)
return dict
}
return strings as! [String: String]
}
func localizedString(_ key: String) -> String {
let specificKey = key + "#SPECIAL"
let tableName = "TableName"
let bundle = Bundle.main
return self.strings(for: tableName)[specificKey]
?? NSLocalizedString(key
, tableName: tableName
, bundle: bundle
, comment: "")
}

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

URL Builder in Swift

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)"

Split URL query in Swift

I have the following URL query:
encodedMessage=PD94bWwgdmVyNlPg%3D%3D&signature=kcig33sdAOAr%2FYYGf5r4HGN
How can I split the query to get the of encodedMessage and signature values?
The right way to achieve this is to work with URLComponents:
A structure designed to parse URLs based on RFC 3986 and to construct
URLs from their constituent parts.
By getting the url components host string and query​Items array, as follows:
if let urlComponents = URLComponents(string: "http://mydummysite.com?encodedMessage=PD94bWwgdmVyNlPg%3D%3D&signature=kcig33sdAOAr%2FYYGf5r4HGN"), let host = urlComponents.host, let queryItems = urlComponents.queryItems {
print(host) // mydummysite.com
print(queryItems) // [encodedMessage=PD94bWwgdmVyNlPg==, signature=kcig33sdAOAr/YYGf5r4HGN]
}
queryItems array contains URLQuery​Item objects, which have name and value properties:
if let urlComponents = URLComponents(string: "http://mydummysite.com?encodedMessage=PD94bWwgdmVyNlPg%3D%3D&signature=kcig33sdAOAr%2FYYGf5r4HGN"),let queryItems = urlComponents.queryItems {
// for example, we will get the first item name and value:
let name = queryItems[0].name // encodedMessage
let value = queryItems[0].value // PD94bWwgdmVyNlPg==
}
Also:
In case of you are getting the query without the full url, I'd suggest to do a pretty simple trick, by adding a dummy host as a prefix to your query string, as follows:
let myQuery = "encodedMessage=PD94bWwgdmVyNlPg%3D%3D&signature=kcig33sdAOAr%2FYYGf5r4HGN"
let myDummyUrlString = "http://stackoverflow.com?" + myQuery
if let urlComponents = URLComponents(string: myDummyUrlString),let queryItems = urlComponents.queryItems {
// for example, we will get the first item name and value:
let name = queryItems[0].name // encodedMessage
let value = queryItems[0].value // PD94bWwgdmVyNlPg==
} else {
print("invalid url")
}
You can get the key value pairs this way:
let str = "encodedMessage=PD94bWwgdmVyNlPg%3D%3D&signature=kcig33sdAOAr%2FYYGf5r4HGN"
let arr = str.components(separatedBy:"&")
var data = [String:Any]()
for row in arr {
let pairs = row.components(separatedBy:"=")
data[pairs[0]] = pairs[1]
}
let message = data["encodedMessage"]
let sig = data["signature"]
I am not sure if that's what you were looking for or not. If it is not, could you please clarify a bit further as to what you are looking for?

How to get the video ID of a YouTube string

(XCode 6.3.2, Swift 1.2)
After researching the internet for the whole evening I already know that this can't be done that easily.
I simply want to get the video ID of a YouTube link. So the "ABCDE" in "https://www.youtube.com/watch?v=ABCDE"
What I got so far:
var url = "https://www.youtube.com/watch?v=ABCDE"
let characterToFind: Character = "="
let characterIndex = find(url, characterToFind)
println(characterIndex)
url.substringFromIndex(advance(url.startIndex, characterIndex))
Prinln outputs: Optional(31)
That's right but I can't use this because it's no index.
XCode also states for the last line: Missing argument for parameter #3 in call
I also don't know what the 3rd parameter of substringFromIndex should be.
Many thanks in advance!
In your case there is no need to create an NSURL as other answers do. Just use:
var url = "https://www.youtube.com/watch?v=ABCDE"
if let videoID = url.componentsSeparatedByString("=").last {
print(videoID)
}
Swift 3+ version:
var url = "https://www.youtube.com/watch?v=ABCDE"
if let videoID = url.components(separatedBy: "=").last {
print(videoID)
}
You can use NSURL query property as follow:
Xcode 8.3.1 • Swift 3.1
let link = "https://www.youtube.com/watch?v=ABCDE"
if let videoID = URL(string: link)?.query?.components(separatedBy: "=").last {
print(videoID) // "ABCDE"
}
Another option is to use URLComponents:
let link = "https://www.youtube.com/watch?v=ABCDE"
if let videoID = URLComponents(string: link)?.queryItems?.filter({$0.name == "v"}).first?.value {
print(videoID) // "ABCDE"
}
Swift 1.x
let link = "https://www.youtube.com/watch?v=ABCDE"
if let videoID = NSURL(string: link)?.query?.componentsSeparatedByString("=").last {
println(videoID) // "ABCDE"
}
You need to unwrap the optional to use the index:
var url = "https://www.youtube.com/watch?v=ABCDE"
let characterToFind: Character = "="
if let index = find(url, characterToFind) { // Unwrap the optional
url.substringFromIndex(advance(index, 1)) // => "ABCDE"
}
find returns an optional – because the character might not be found, in which case it will be nil. You need to unwrap it to check it isn’t and to get the actual value out.
You can also use a range-based subscript on the index directly, rather than using advance to turn it into an integer index:
let url = "https://www.youtube.com/watch?v=ABCDE"
if let let characterIndex = find(url, "=") {
let value = url[characterIndex.successor()..<url.endIndex] // ABCDE
}
else {
// error handling, if you want it, here
}
You have more options if there is a reasonable default in the case of “not found” For example, if you just want an empty string in that case, this will work:
let idx = find(url, "=")?.successor() ?? url.endIndex
let value = url[idx..<url.endIndex]
Or, maybe you don’t even need to deal with the optionality right now, so you’re happy to leave the result optional as well:
// value will be Optional("ABCD")
let value = find(url, "=").map { url[$0.successor()..<url.endIndex] }
For a rundown on optionals, take a look here and here.
Also, rather than hand-parsing URL strings at all, you might find the info here useful.
With your url format, you can get the 'v' parameter's value by converting to NSURL then get parameters and separate it by '='.
var url: NSURL = NSURL(string: "https://www.youtube.com/watch?v=RAzOOfVA2_8")!
url.query?.componentsSeparatedByString("=").last
url.query returns v=RAzOOfVA2_8
If the link has more than 1 parameter, you can get all parameters then do a loop to verify each parameter:
var url: NSURL = NSURL(string: "https://www.youtube.com/watch?v=RAzOOfVA2_8")!
var params = url.query?.componentsSeparatedByString("&")
if let _params = params { //have parameters in URL
for param in _params { //verify each pair of key & value in your parameters
var _paramArray = param.componentsSeparatedByString("=") //separate by '=' to get key & value
if (_paramArray.first?.compare("v", options: nil, range: nil, locale: nil) == NSComparisonResult.OrderedSame) {
println(_paramArray.last)
}
}
} else {
//url does not have any parameter
}