Objective-c编码字符串?

我发现两者 stringByAddingPercentEscapesUsingEncoding: 和 CFURLCreateStringByAddingPercentEscapes() 这是不够的。这个 NSString

为了解决这个问题,我创建了一个 NSString字符串 category方法来正确编码字符串。它会把所有的东西 [a-zA-Z0-9.-_~] + (根据 this specification ). 它还将正确处理编码unicode字符。

- (NSString *) URLEncodedString_ch {
    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;
}

iOS7SDK现在有了一个更好的替代方案 stringByAddingPercentEscapesUsingEncoding 这允许您指定要转义除某些允许的字符外的所有字符。如果您将URL分为以下几个部分进行构建,则效果很好:

NSString * unescapedQuery = [[NSString alloc] initWithFormat:@"?myparam=%d", numericParamValue];
NSString * escapedQuery = [unescapedQuery stringByAddingPercentEncodingWithAllowedCharacters:[NSCharacterSet URLQueryAllowedCharacterSet]];
NSString * urlString = [[NSString alloc] initWithFormat:@"http://ExampleOnly.com/path.ext%@", escapedQuery];

[NSCharacterSet URLHostAllowedCharacterSet]
[NSCharacterSet URLUserAllowedCharacterSet]
[NSCharacterSet URLPasswordAllowedCharacterSet]
[NSCharacterSet URLPathAllowedCharacterSet]
[NSCharacterSet URLFragmentAllowedCharacterSet]

[NSCharacterSet URLQueryAllowedCharacterSet] 包括 全部的 URL的查询部分(以 ? # 对于碎片,如果有的话)包括 ? & 或 = 零件

NSMutableCharacterSet * URLQueryPartAllowedCharacterSet; // possibly defined in class extension ...

// ... and built in init or on first use
URLQueryPartAllowedCharacterSet = [[NSCharacterSet URLQueryAllowedCharacterSet] mutableCopy];
[URLQueryPartAllowedCharacterSet removeCharactersInString:@"&+=?"]; // %26, %3D, %3F

// then escape variables in the URL, such as values in the query and any fragment:
NSString * escapedValue = [anUnescapedValue stringByAddingPercentEncodingWithAllowedCharacters:URLQueryPartAllowedCharacterSet];
NSString * escapedFrag = [anUnescapedFrag stringByAddingPercentEncodingWithAllowedCharacters:[NSCharacterSet URLFragmentAllowedCharacterSet]];
NSString * urlString = [[NSString alloc] initWithFormat:@"http://ExampleOnly.com/path.ext?myparam=%@#%@", escapedValue, escapedFrag];
NSURL * url = [[NSURL alloc] initWithString:urlString];

unescapedValue 甚至可以是整个URL,例如回调或重定向:

NSString * escapedCallbackParamValue = [anAlreadyEscapedCallbackURL stringByAddingPercentEncodingWithAllowedCharacters:URLQueryPartAllowedCharacterSet];
NSURL * callbackURL = [[NSURL alloc] initWithString:[[NSString alloc] initWithFormat:@"http://ExampleOnly.com/path.ext?callback=%@", escapedCallbackParamValue]];

NSURL initWithScheme:(NSString *)scheme host:(NSString *)host path:(NSString *)path 对于带有查询字符串的URL,因为它将向路径添加更多的转义百分比。

NSString *encodedString = [myString stringByAddingPercentEscapesUsingEncoding:NSASCIIStringEncoding];

注意--新字符串将 autorelease

NSString的 stringByAddingPercentEscapesUsingEncoding: 看起来像你要找的。

编辑 CFURLCreateStringByAddingPercentEscapes 相反。 originalString 可以是 NSString 或者 CFStringRef

CFStringRef newString = CFURLCreateStringByAddingPercentEscapes(kCFAllocatorDefault, originalString, NULL, CFSTR("!*'();:@&=+@,/?#[]"), kCFStringEncodingUTF8);

请注意,这是未经测试。你应该看一看 documentation page ,免费搭桥的想法,等等。

另外,我也不知道(在我的脑子里)在 legalURLCharactersToBeEscaped

我把这个答案做成一个社区wiki,这样对CoreFoundation有更多了解的人就可以做出改进。

// https://tools.ietf.org/html/rfc3986#section-2.2
let rfc3986Reserved = NSCharacterSet(charactersInString: "!*'();:@&=+$,/?#[]")
let encoded = "email+with+plus@example.com".stringByAddingPercentEncodingWithAllowedCharacters(rfc3986Reserved.invertedSet)

输出: email%2Bwith%2Bplus%40example.com

ASI HttpRequest library 在objective-c程序中(我不能高度推荐),您可以在其ASIFormDataRequest对象上使用“encodeURL”助手API。不幸的是,API不是静态的,所以在您的项目中使用它的实现创建一个扩展可能是值得的。

- (NSString*)encodeURL:(NSString *)string
{
    NSString *newString = NSMakeCollectable([(NSString *)CFURLCreateStringByAddingPercentEscapes(kCFAllocatorDefault, (CFStringRef)string, NULL, CFSTR(":/?#[]@!$ &'()*+,;=\"<>%{}|\\^~`"), CFStringConvertNSStringEncodingToEncoding([self stringEncoding])) autorelease]);
    if (newString) {
        return newString;
    }
    return @"";
}

如您所见,它本质上是一个包装器 CFURLCreateStringByAddingPercentEscapes

在我注意到Rob的答案之前,我把Dave的答案传给了Swift,这个答案似乎很有效,而且因为它更干净而更受欢迎。如果有人感兴趣,我就把它留在这里:

public extension String {

    // For performance, I've replaced the char constants with integers, as char constants don't work in Swift.

    var URLEncodedValue: String {
        let output = NSMutableString()
        guard let source = self.cStringUsingEncoding(NSUTF8StringEncoding) else {
            return self
        }
        let sourceLen = source.count

        var i = 0
        while i < sourceLen - 1 {
            let thisChar = source[i]
            if thisChar == 32 {
                output.appendString("+")
            } else if thisChar == 46 || thisChar == 45 || thisChar == 95 || thisChar == 126 ||
                (thisChar >= 97 && thisChar <= 122) ||
                (thisChar >= 65 && thisChar <= 90) ||
                (thisChar >= 48 && thisChar <= 57) {
                    output.appendFormat("%c", thisChar)
            } else {
                output.appendFormat("%%%02X", thisChar)
            }

            i++
        }

        return output as String
    }
}

 var str = "someurl/somecontent"

 let percentEncodedString = str.addingPercentEncoding(withAllowedCharacters: .alphanumerics)