How do I URL encode a string

2018-12-31 09:01发布

I have a URL string (NSString) with spaces and & characters. How do I url encode the entire string (including the & ampersand character and spaces)?

21条回答
步步皆殇っ
2楼-- · 2018-12-31 09:32

Try to use stringByAddingPercentEncodingWithAllowedCharacters method with [NSCharacterSet URLUserAllowedCharacterSet] it will cover all the cases

Objective C

NSString *value = @"Test / Test";
value = [value stringByAddingPercentEncodingWithAllowedCharacters:[NSCharacterSet URLUserAllowedCharacterSet]];

swift

var value = "Test / Test"
value.stringByAddingPercentEncodingWithAllowedCharacters(NSCharacterSet.URLUserAllowedCharacterSet())

Output

Test%20%2F%20Test

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笑指拈花
3楼-- · 2018-12-31 09:32
-(NSString *)encodeUrlString:(NSString *)string {
  return CFBridgingRelease(
                    CFURLCreateStringByAddingPercentEscapes(
                        kCFAllocatorDefault,
                        (__bridge CFStringRef)string,
                        NULL,
                        CFSTR("!*'();:@&=+$,/?%#[]"),
                        kCFStringEncodingUTF8)
                    );
}

according to the following blog

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看淡一切
4楼-- · 2018-12-31 09:32

This one is working for me.

func stringByAddingPercentEncodingForFormData(plusForSpace: Bool=false) -> String? {
    let unreserved = "*-._"
    let allowed = NSMutableCharacterSet.alphanumericCharacterSet()
    allowed.addCharactersInString(unreserved)

    if plusForSpace {
        allowed.addCharactersInString(" ")
    }

    var encoded = stringByAddingPercentEncodingWithAllowedCharacters(allowed)
    if plusForSpace {
        encoded = encoded?.stringByReplacingOccurrencesOfString(" ",
                                                                withString: "+")
    }
    return encoded
}

I found the above function from this link: http://useyourloaf.com/blog/how-to-percent-encode-a-url-string/

You can also use this function with swift extension. Please let me know if there is any issue.

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萌妹纸的霸气范
5楼-- · 2018-12-31 09:33

In Swift 3, please try out below:

let stringURL = "YOUR URL TO BE ENCODE";
let encodedURLString = stringURL.addingPercentEncoding(withAllowedCharacters: .urlHostAllowed)
print(encodedURLString)

Since, stringByAddingPercentEscapesUsingEncoding encodes non URL characters but leaves the reserved characters (like !*'();:@&=+$,/?%#[]), You can encode the url like the following code:

let stringURL = "YOUR URL TO BE ENCODE";
let characterSetTobeAllowed = (CharacterSet(charactersIn: "!*'();:@&=+$,/?%#[] ").inverted)
if let encodedURLString = stringURL.addingPercentEncoding(withAllowedCharacters: characterSetTobeAllowed) {
    print(encodedURLString)
}
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何处买醉
6楼-- · 2018-12-31 09:34

I opted to use the CFURLCreateStringByAddingPercentEscapes call as given by accepted answer, however in newest version of XCode (and IOS), it resulted in an error, so used the following instead:

NSString *apiKeyRaw = @"79b|7Qd.jW=])(fv|M&W0O|3CENnrbNh4}2E|-)J*BCjCMrWy%dSfGs#A6N38Fo~";

NSString *apiKey = (NSString *)CFBridgingRelease(CFURLCreateStringByAddingPercentEscapes(NULL, (CFStringRef)apiKeyRaw, NULL, (CFStringRef)@"!*'();:@&=+$,/?%#[]", kCFStringEncodingUTF8));
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大哥的爱人
7楼-- · 2018-12-31 09:36

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#15674 http://simonwoodside.com/weblog/2009/4/22/how_to_really_url_encode/

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