How do I URL encode a string

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#15674http://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)]; // %26encodedString = [myString stringByAddingPercentEncodingWithAllowedCharacters:chars];