关于 iOS 不受支持的 URL 问题

https://blog.csdn.net/huanghuipost/article/details/102521074

https://www.cnblogs.com/thinksasa/p/3530813.html

https://www.cnblogs.com/huzi007/p/4174519.html

https://blog.csdn.net/weixin_33901926/article/details/89039096

https://www.jianshu.com/p/dbb15f019127?utm_campaign=maleskine&utm_content=note&utm_medium=seo_notes&utm_source=recommendation

 

第一：

URL中含有转义字符例如   双引号 ""  、大括号{}等

 

 

用stringByAddingPercentEncodingWithAllowedCharacters取代CFURLCreateStringByAddingPercentEscapes

1.网络访问请求：中文空格字符编码/解码

在 iOS 程序访问 HTTP 资源时需要对 URL 进行 Encode，
比如像拼出来的 http://unmi.cc?p1=%+&sd f&p2=中文，
其中的中文、特殊符号&％和空格都必须进行转译才能正确访问。
现在以“?!@#$^&%*+,:;='\”`<>()[]{}/\| "字符串为例子，
用stringByAddingPercentEncodingWithAllowedCharacters取代
CFURLCreateStringByAddingPercentEscapes

stringByAddingPercentEscapesUsingEncoding
(只对 `#%^{}[]|"<> 加空格共14个字符编码，不包括“&?”等符号)，
iOS 9将淘汰，建议用stringByAddingPercentEncodingWithAllowedCharacters方法

URLUserAllowedCharacterSet      "#%/:<>?@[\]^`

URLPasswordAllowedCharacterSet  "#%/:<>?@[\]^`{|}

URLHostAllowedCharacterSet      "#%/<>?@\^`{|}

URLPathAllowedCharacterSet      "#%;<>?[\]^`{|}

URLQueryAllowedCharacterSet     "#%<>[\]^`{|}

URLFragmentAllowedCharacterSet  "#%<>[\]^`{|}

2.网络访问请求：中文空格字符解码
stringByRemovingPercentEncoding :
Xcode7可能会提示要将stringByAddingPercentEscapesUsingEncoding替换成此方法，要根据是否是解码来区分

//代替stringByAddingPercentEscapesUsingEncoding 
let customAllowedSet = NSCharacterSet(charactersInString:"`#%^{}\"[]|\\<> ").invertedSet

NSString *resourcePath = @"http://www.jianshu.com/users/ac47ba96ae2f/latest_articles";     
NSString *encodePath ;       
if (!IOS7_OR_LATER) {
        encodePath = [resourcePath stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding];    
} else {
        encodePath = [resourcePath stringByAddingPercentEncodingWithAllowedCharacters:[NSCharacterSet characterSetWithCharactersInString:@"`#%^{}\"[]|\\<> "].invertedSet];    
}

举例：

原字符串为：

NSString *url = @"ertehtt""p://xxdsdscrg?!@#$^&%*+,:;='\"`<>()[]{}/\\| ";

CFURLCreateStringByAddingPercentEscapes方法为：

CFStringRef encodedCFString = CFURLCreateStringByAddingPercentEscapes(kCFAllocatorDefault, (__bridge CFStringRef) url, nil, CFSTR("?!@#$^&%*+,:;='\"`<>()[]{}/\\| "), kCFStringEncodingUTF8);  
NSString *encodedString = [[NSString alloc] initWithString:(__bridge_transfer NSString*) encodedCFString];

stringByAddingPercentEncodingWithAllowedCharacters方法为：

NSString *charactersToEscape = @"?!@#$^&%*+,:;='\"`<>()[]{}/\\| ";  
NSCharacterSet *allowedCharacters = [[NSCharacterSet characterSetWithCharactersInString:charactersToEscape] invertedSet];  
NSString *encodedUrl = [url stringByAddingPercentEncodingWithAllowedCharacters:allowedCharacters];  
NSLog(@"\n%@\n%@",encodedUrl,encodedString);  

之后获得的字符串为一致的。

问题实例：

我有一个有效的 url，我得到 不受支持的 url 错误。正如你可以看到 http://

//http://www.jianshu.com/users/ac47ba96ae2f/latest_articles
Error description=Error Domain=NSURLErrorDomain Code=-1002 "unsupported URL" UserInfo=0x78f97920 {NSUnderlyingError=0x79f78bd0 "unsupported URL", NSLocalizedDescription=unsupported URL}

这是我想试图如何初始化 url:
方法 1:

NSString *path = @"http://www.jianshu.com/users/ac47ba96ae2f/latest_articles";
NSURL *url = [NSURL URLWithString:path];
NSMutableURLRequest *request =[NSMutableURLRequest requestWithURL:url];

方法 2:

NSMutableURLRequest *request =[NSMutableURLRequest requestWithURL:[NSURL URLWithString:@"http://www.jianshu.com/users/ac47ba96ae2f/latest_articles"]];

解决方法 :

URL 不能包含 ASCII 字符集中, 不是必须这样的字符进行转义的字符。
使用 stringByAddingPercentEncodingWithAllowedCharacters
字符集 URLQueryAllowedCharacterSet

NSString *path = @"http://www.jianshu.com/users/ac47ba96ae2f/latest_articles";
NSString *escapedPath = [path stringByAddingPercentEncodingWithAllowedCharacters:[NSCharacterSet URLQueryAllowedCharacterSet]];
NSLog(@"escapedPath: %@", escapedPath);

输出:

escapedPath: http://www.jianshu.com/users/ac47ba96ae2f/latest_articles

作者：Kingsleeeey
链接：https://www.jianshu.com/p/dbb15f019127
来源：简书
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