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Editorial Team
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Asked: 2026-05-26T20:45:49+00:00

I have an NSString like this: http://www. But I want to transform it to

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I have an NSString like this:

http://www.

But I want to transform it to this:

http%3A%2F%2Fwww.

How can I do this?

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1 Answer

  1. Editorial Team

    To escape the characters you want is a little more work.

    Example code

    iOS7 and above:

    NSString *unescaped = @"http://www";
NSString *escapedString = [unescaped stringByAddingPercentEncodingWithAllowedCharacters:[NSCharacterSet URLHostAllowedCharacterSet]];
NSLog(@"escapedString: %@", escapedString);

    NSLog output:

    escapedString: http%3A%2F%2Fwww

    The following are useful URL encoding character sets:

    URLFragmentAllowedCharacterSet  "#%<>[\]^`{|}
URLHostAllowedCharacterSet      "#%/<>?@\^`{|}
URLPasswordAllowedCharacterSet  "#%/:<>?@[\]^`{|}
URLPathAllowedCharacterSet      "#%;<>?[\]^`{|}
URLQueryAllowedCharacterSet     "#%<>[\]^`{|}
URLUserAllowedCharacterSet      "#%/:<>?@[\]^`

    Creating a characterset combining all of the above:

    NSCharacterSet *URLCombinedCharacterSet = [[NSCharacterSet characterSetWithCharactersInString:@" \"#%/:<>?@[\\]^`{|}"] invertedSet];

    Creating a Base64

    In the case of Base64 characterset: 

    NSCharacterSet *URLBase64CharacterSet = [[NSCharacterSet characterSetWithCharactersInString:@"/+=\n"] invertedSet];

    For Swift 3.0:

    var escapedString = originalString.addingPercentEncoding(withAllowedCharacters:.urlHostAllowed)

    For Swift 2.x:

    var escapedString = originalString.stringByAddingPercentEncodingWithAllowedCharacters(NSCharacterSet.URLHostAllowedCharacterSet())

    Note: stringByAddingPercentEncodingWithAllowedCharacters will also encode UTF-8 characters needing encoding.

    Pre iOS7 use Core Foundation
    Using Core Foundation With ARC: 

    NSString *escapedString = (NSString *)CFBridgingRelease(CFURLCreateStringByAddingPercentEscapes(
    NULL,
   (__bridge CFStringRef) unescaped,
    NULL,
    CFSTR("!*'();:@&=+$,/?%#[]\" "),
    kCFStringEncodingUTF8));

    Using Core Foundation Without ARC: 

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

    Note: -stringByAddingPercentEscapesUsingEncoding will not produce the correct encoding, in this case it will not encode anything returning the same string.

    stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding encodes 14 characrters:

    `#%^{}[]|\”<> plus the space character as percent escaped.

    testString: 

    " `~!@#$%^&*()_+-={}[]|\\:;\"'<,>.?/AZaz"

    encodedString: 

    "%20%60~!@%23$%25%5E&*()_+-=%7B%7D%5B%5D%7C%5C:;%22'%3C,%3E.?/AZaz"

    Note: consider if this set of characters meet your needs, if not change them as needed.

    RFC 3986 characters requiring encoding (% added since it is the encoding prefix character):

    “!#$&'()*+,/:;=?@[]%”

    Some “unreserved characters” are additionally encoded:

    “\n\r \”%-.<>\^_`{|}~”

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