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?

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