CODEWITHSUNDEEP
javascriptobjective-c
Trung
Trung

Reputation: 155

encodeURIComponent equivalent object-c

I'm having trouble googling this.

Does Objective-C have an equivalent method for encoding URI Component?

http://www.w3schools.com/jsref/jsref_encodeuricomponent.asp

This seems to be a very common use case but I can't find any documentation about it in Objective-C.

Thanks.

Upvotes: 4

Views: 4077

Answers (4)

Burcea Bogdan Madalin
Burcea Bogdan Madalin

Reputation: 194

Using CFURLCreateStringByAddingPercentEscapes or stringByAddingPercentEscapesUsingEncoding is not ideal since those are deprecated methods.

Dave's answer is on the right track but is not explicit enough.

We want to call stringByAddingPercentEncodingWithAllowedCharacters but we need to choose a character set. Despite what our intuiton might say, NSCharacterSet.URLQueryAllowedCharacterSet is actually not a good choice. It refers to the whole query part of the URL, not just to values of query parameters so it allows characters like ? or & as well, which are not valid here. What you want is to manually build the allowed characters set according to the documentation and use that. The allowed chars for encodeURIComponent() are:

A–Z a–z 0–9 - _ . ! ~ * ' ( )

That gives us this solution:

// Note: Using NSCharacterSet.URLQueryAllowedCharacterSet is inappropriate for this purpose.
// It contains the allowed chars for the whole Query part of the URL (including '?' or '&'),
// not just for the values of query parameters (where chars are more restricted).

+ (NSString *)encodeURIComponent:(NSString *)uriComponent {
    // https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/encodeURIComponent
    NSString *allowedChars = @"ABCDEFGHIJKLMNOPQRSTUVWXYZ"
                              "abcdefghijklmnopqrstuvwxyz"
                              "0123456789"
                              "-_.!~*'()";

    NSCharacterSet *charSet = [NSCharacterSet characterSetWithCharactersInString:allowedChars];
    return [uriComponent stringByAddingPercentEncodingWithAllowedCharacters:charSet];
}

+ (NSString *)decodeURIComponent:(NSString *)uriComponent {
    return [uriComponent stringByRemovingPercentEncoding];
}

Upvotes: 0

bendytree
bendytree

Reputation: 13629

Now stringByAddingPercentEscapesUsingEncoding: is deprecated. Here's the iOS 7+ equivalent of encodeURIComponent:

[str stringByAddingPercentEncodingWithAllowedCharacters:[NSCharacterSet URLQueryAllowedCharacterSet]]

Upvotes: 0

HiroGuo
HiroGuo

Reputation: 21

off course, you can choose this methods, I hope that will be helpful.

//encode URL string

+(NSString *)URLEncodedString:(NSString *)str
{

    NSString *encodedString = (NSString *)
    CFBridgingRelease(CFURLCreateStringByAddingPercentEscapes(kCFAllocatorDefault,
                                                              (CFStringRef)str,
                                                              NULL,
                                                              (CFStringRef)@"!*'();:@&=+$,/?%#[]",
                                                              kCFStringEncodingUTF8));

    return encodedString;
}

}

//decode URL string

+(NSString *)URLDecodedString:(NSString *)str
{

  NSString *decodedString=(__bridge_transfer NSString *)CFURLCreateStringByReplacingPercentEscapesUsingEncoding(NULL, (__bridge CFStringRef)str, CFSTR(""), CFStringConvertNSStringEncodingToEncoding(NSUTF8StringEncoding));

    return decodedString;
}

Upvotes: 1

Dave DeLong
Dave DeLong

Reputation: 243156

-[NSString stringByAddingPercentEscapesUsingEncoding:] is the simplest way to do this.

Upvotes: 6

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