How do I URL encode a string

Question

Unfortunately, stringByAddingPercentEscapesUsingEncoding doesn't always work 100%. It encodes non-URL characters but leaves the reserved characters (like slash / and ampersand &) alone. Apparently this is a bug that Apple is aware of, but since they have not fixed it yet, I have been using this category to url-encode a string:

@implementation NSString (NSString_Extended)

- (NSString *)urlencode {
    NSMutableString *output = [NSMutableString string];
    const unsigned char *source = (const unsigned char *)[self UTF8String];
    int sourceLen = strlen((const char *)source);
    for (int i = 0; i < sourceLen; ++i) {
        const unsigned char thisChar = source[i];
        if (thisChar == ' '){
            [output appendString:@"+"];
        } else if (thisChar == '.' || thisChar == '-' || thisChar == '_' || thisChar == '~' || 
                   (thisChar >= 'a' && thisChar <= 'z') ||
                   (thisChar >= 'A' && thisChar <= 'Z') ||
                   (thisChar >= '0' && thisChar <= '9')) {
            [output appendFormat:@"%c", thisChar];
        } else {
            [output appendFormat:@"%%%02X", thisChar];
        }
    }
    return output;
}

Used like this:

NSString *urlEncodedString = [@"SOME_URL_GOES_HERE" urlencode];

// Or, with an already existing string:
NSString *someUrlString = @"someURL";
NSString *encodedUrlStr = [someUrlString urlencode];

---

This also works:

NSString *encodedString = (NSString *)CFURLCreateStringByAddingPercentEscapes(
                            NULL,
                            (CFStringRef)unencodedString,
                            NULL,
                            (CFStringRef)@"!*'();:@&=+$,/?%#[]",
                            kCFStringEncodingUTF8 );

---

Some good reading about the subject:

Objective-c iPhone percent encode a string?
Objective-C and Swift URL encoding

http://cybersam.com/programming/proper-url-percent-encoding-in-ios
https://devforums.apple.com/message/15674#15674 http://simonwoodside.com/weblog/2009/4/22/how_to_really_url_encode/

This might be helpful

NSString *sampleUrl = @"http://www.google.com/search.jsp?params=Java Developer";
NSString* encodedUrl = [sampleUrl stringByAddingPercentEscapesUsingEncoding:
 NSUTF8StringEncoding];

For iOS 7+, the recommended way is:

NSString* encodedUrl = [sampleUrl stringByAddingPercentEncodingWithAllowedCharacters:[NSCharacterSet URLQueryAllowedCharacterSet]];

You can choose the allowed character set as per the requirement of the URL component.

New APIs have been added since the answer was selected; You can now use NSURLUtilities. Since different parts of URLs allow different characters, use the applicable character set. The following example encodes for inclusion in the query string:

encodedString = [myString stringByAddingPercentEncodingWithAllowedCharacters:NSCharacterSet.URLQueryAllowedCharacterSet];

To specifically convert '&', you'll need to remove it from the url query set or use a different set, as '&' is allowed in a URL query:

NSMutableCharacterSet *chars = NSCharacterSet.URLQueryAllowedCharacterSet.mutableCopy;
[chars removeCharactersInRange:NSMakeRange('&', 1)]; // %26
encodedString = [myString stringByAddingPercentEncodingWithAllowedCharacters:chars];

Swift 2.0 Example (iOS 9 Compatiable)

extension String {

  func stringByURLEncoding() -> String? {

    let characters = NSCharacterSet.URLQueryAllowedCharacterSet().mutableCopy() as! NSMutableCharacterSet

    characters.removeCharactersInString("&")

    guard let encodedString = self.stringByAddingPercentEncodingWithAllowedCharacters(characters) else {
      return nil
    }

    return encodedString

  }

}