Objective-C and Swift URL encoding

Question

I have a NSString like this:

http://www.

but I want to transform it to:

http%3A%2F%2Fwww.

How can I do this?

Solution 1

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 \"%-.<>\^_`{|}~"

Solution 2

It's called URL encoding. More here.

-(NSString *)urlEncodeUsingEncoding:(NSStringEncoding)encoding {
    return (NSString *)CFURLCreateStringByAddingPercentEscapes(NULL,
           (CFStringRef)self,
           NULL,
           (CFStringRef)@"!*'\"();:@&=+$,/?%#[]% ",
           CFStringConvertNSStringEncodingToEncoding(encoding));
}

Solution 3

This is not my solution. Someone else wrote in stackoverflow but I have forgotten how.

Somehow this solution works "well". It handles diacritic, chinese characters, and pretty much anything else.

- (NSString *) URLEncodedString {
    NSMutableString * output = [NSMutableString string];
    const char * source = [self UTF8String];
    int sourceLen = strlen(source);
    for (int i = 0; i < sourceLen; ++i) {
        const unsigned char thisChar = (const unsigned char)source[i];
        if (false && 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;
}

If someone would tell me who wrote this code, I'll really appreciate it. Basically he has some explanation why this encoded string will decode exactly as it wish.

I modified his solution a little. I like space to be represented with %20 rather than +. That's all.

Solution 4

 NSString * encodedString = (NSString *)CFURLCreateStringByAddingPercentEscapes(NUL,(CFStringRef)@"parameter",NULL,(CFStringRef)@"!*'();@&+$,/?%#[]~=_-.:",kCFStringEncodingUTF8 );

NSURL * url = [[NSURL alloc] initWithString:[@"address here" stringByAppendingFormat:@"?cid=%@",encodedString, nil]];

Solution 5

This can work in Objective C ARC.Use CFBridgingRelease to cast a Core Foundation-style object as an Objective-C object and transfer ownership of the object to ARC .See Function CFBridgingRelease here.

+ (NSString *)encodeUrlString:(NSString *)string {
return CFBridgingRelease(CFURLCreateStringByAddingPercentEscapes
                         (kCFAllocatorDefault,
                          (__bridge CFStringRef)string,
                          NULL,
                          CFSTR("!*'();:@&=+$,/?%#[]"),
                          kCFStringEncodingUTF8)
                         );}

View more solutions

Comments

[chown]

This would be a lot more useful if some content from the links you posted were included in the answer.

[Mike Weller]

Also note you can use NSString's -stringByAddingPercentEscapesUsingEncoding method.

[Mike Weller]

This code also leaks memory unless you use a __bridge_retained cast under ARC, or call CFRelease at some point on the return value from CFURLCreateStringByAddingPercentEscapes.

[Mike Weller]

Ah yes, now I remember the funky stringByAddingPercentEscapesUsingEncoding behaviour. It only encodes '&' and '=' or something ridiculous like that.

[zaph]

Under ARC the initial answer does leak and the suggestion by @Mike is correct, I have added alternatives for use under ARC.

[Praveen]

also under ARC you would need to set the bridge for the unescaped string: ... (__bridge CFStringRef) unescaped, ...

[Zahi]

release encodedString and url. this code is about encode parameter. To encode whole address pass string instead of "parameter".

[Sébastien Stormacq]

stringByReplacingPercentEscapeusingencoding: only escapes & and = :-(

[Alex Nauda]

According to RFC1738 you would need to encode additional characters as well. So although this does answer the OP's question, it has limited usefulness as a general-purpose URL encoder. For example, it doesn't handle non-alphanumerics such as a German umlaut.

[John Ballinger]

Correct, so things like + are not encoded, when they need to be. So dont use the above answer

[lgdev]

Maybe for more usability, should the solution contain the space and quote characters? Something like CFSTR("!*'();:@&=+$,/?%#[]\" ")

[zaph]

@Luka Updated answer.

[hsoi]

Original code: stackoverflow.com/questions/3423545/…

[coolcool1994]

This doesn't work (for iOS 7 part). This doesn't convert & into %26.

[zaph]

Create the character set you need from an NSString with characterSetWithCharactersInString, take the inverse with invertedSet and use that with stringByAddingPercentEncodingWithAllowedCharacters. For an example see this SO answer.

[lifeisfoo]

The Apple documentation says: Important: Although the NSString class provides built-in methods for adding percent escapes, you usually should not use them.

[Виктор Иванов]

This worked for me... encoded even the & character that I was having issues with.

[Yuchao Zhou]

what is [self UTF8String] ?

[Kelsey]

@yuchaozh this is a function that you put inside a category of NSString. So, self is NSStirng and [self UTF8String] returns, assumingly, its string in UTF8 format.

[Pang]

CFURLCreateStringByAddingPercentEscapes() is deprecated. Use [NSString stringByAddingPercentEncodingWithAllowedCharacters:] instead.