Objective-C 和 Swift URL 编码 [英] Objective-C and Swift URL encoding

问题描述

我有一个这样的 NSString:

http://www.

但我想将其转换为:

http%3A%2F%2Fwww.

我该怎么做?

解决方案

要转义您想要的字符需要更多的工作.

示例代码

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iOS7 及以上:

NSString *unescaped = @"http://www";NSString *escapedString = [未转义的 stringByAddingPercentEncodingWithAllowedCharacters:[NSCharacterSet URLHostAllowedCharacterSet]];NSLog(@"escapedString: %@", escapedString);

NSLog 输出:

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转义字符串:http%3A%2F%2Fwww

以下是有用的 URL 编码字符集:

URLFragmentAllowedCharacterSet "#%<>[]^`{|}URLHostAllowedCharacterSet "#%/<>?@^`{|}URLPasswordAllowedCharacterSet "#%/:<>?@[]^`{|}URLPathAllowedCharacterSet "#%;<>?[]^`{|}URLQueryAllowedCharacterSet "#%<>[]^`{|}URLUserAllowedCharacterSet "#%/:<>?@[]^`

创建一个结合以上所有内容的字符集:

NSCharacterSet *URLCombinedCharacterSet = [[NSCharacterSet characterSetWithCharactersInString:@" "#%/:<>?@[\]^`{|}"] reverseSet];

创建一个 Base64

以 Base64 字符集为例:

NSCharacterSet *URLBase64CharacterSet = [[NSCharacterSet characterSetWithCharactersInString:@"/+=
"] reverseSet];

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对于 Swift 3.0:

var escapedString = originalString.addingPercentEncoding(withAllowedCharacters:.urlHostAllowed)

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对于 Swift 2.x:

var escapedString = originalString.stringByAddingPercentEncodingWithAllowedCharacters(NSCharacterSet.URLHostAllowedCharacterSet())

注意:stringByAddingPercentEncodingWithAllowedCharacters 也会对需要编码的 UTF-8 字符进行编码.

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iOS7 之前使用 Core Foundation
在 ARC 中使用 Core Foundation:

NSString *escapedString = (NSString *)CFBridgingRelease(CFURLCreateStringByAddingPercentEscapes(空值，(__bridge CFStringRef) 未转义，空值，CFSTR("!*'();:@&=+$,/?%#[]" "),kCFStringEncodingUTF8));

在没有 ARC 的情况下使用 Core Foundation:

NSString *escapedString = (NSString *)CFURLCreateStringByAddingPercentEscapes(空值，(CFStringRef) 未转义，空值，CFSTR("!*'();:@&=+$,/?%#[]" "),kCFStringEncodingUTF8);

注意:-stringByAddingPercentEscapesUsingEncoding 不会产生正确的编码，在这种情况下它不会编码返回相同字符串的任何内容.

stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding 编码 14 个字符:

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`#%^{}[]|"<> 加上空格字符作为百分比转义.

测试字符串:

" `~!@#$%^&*()_+-={}[]|\:;"'<,>.?/AZaz"

编码字符串:

"%20%60~!@%23$%25%5E&*()_+-=%7B%7D%5B%5D%7C%5C:;%22'%3C,%3E.?/AZAZ"

注意:请考虑这组字符是否满足您的需要，如果没有根据需要进行更改.

需要编码的 RFC 3986 字符(添加 % 因为它是编码前缀字符):

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"!#$&'()*+,/:;=?@[]%"

一些非保留字符"被额外编码:

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

I have a NSString like this:

http://www.

but I want to transform it to:

http%3A%2F%2Fwww.

How can I do this?

解决方案

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:@"/+=
"] 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:

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

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