How do I URL encode a string

Question

I have a URL string (NSString) with spaces and & characters. How do I url encode the entire string (including the & ampersand character and spaces)?

Answer 1

ios 7 update

NSString *encode = [string stringByAddingPercentEncodingWithAllowedCharacters:[NSCharacterSet URLQueryAllowedCharacterSet]];

NSString *decode = [encode stringByReplacingPercentEscapesUsingEncoding:NSUTF8StringEncoding];

Answer 2

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.

Answer 3

After reading all the answers for this topic and the (wrong) accepted one, I want to add my contribution.

IF the target is iOS7+, and in 2017 it should since XCode makes really hard to deliver compatibility under iOS8, the best way, thread safe, fast, amd will full UTF-8 support to do this is:

(Objective C code)

@implementation NSString (NSString_urlencoding)

- (NSString *)urlencode {
    static NSMutableCharacterSet *chars = nil;
    static dispatch_once_t pred;

    if (chars)
        return [self stringByAddingPercentEncodingWithAllowedCharacters:chars];

    // to be thread safe
    dispatch_once(&pred, ^{
        chars = NSCharacterSet.URLQueryAllowedCharacterSet.mutableCopy;
        [chars removeCharactersInString:@"!*'();:@&=+$,/?%#[]"];
    });
    return [self stringByAddingPercentEncodingWithAllowedCharacters:chars];
}
@end

This will extend NSString, will exclude RFC forbidden characters, support UTF-8 characters, and let you use things like:

NSString *myusername = "I'm[evil]&want(to)break!!!$->àéìòù";
NSLog(@"Source: %@ -> Dest: %@", myusername, [myusername urlencode]);

That will print on your debug console:

Source: I'm[evil]&want(to)break!!!$->àéìòù -> Dest: I%27m%5Bevil%5D%26want%28to%29break%21%21%21%24-%3E%C3%A0%C3%A9%C3%AC%C3%B2%C3%B9

... note also the use of dispatch_once to avoid multiple initializations in multithread environments.

Answer 4

I faced a similar problem passing complex strings as a POST parameter. My strings can contain Asian characters, spaces, quotes and all sorts of special characters. The solution I eventually found was to convert my string into the matching series of unicodes, e.g. "Hu0040Hu0020Hu03f5...." using [NSString stringWithFormat:@"Hu%04x",[string characterAtIndex:i]] to get the Unicode from each character in the original string. The same can be done in Java.

This string can be safely passed as a POST parameter.

On the server side (PHP), I change all the "H" to "\" and I pass the resulting string to json_decode. Final step is to escape single quotes before storing the string into MySQL.

This way I can store any UTF8 string on my server.

Answer 5

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];

Answer 6

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

  }

}