關於 iOS 不受支持的 URL 問題

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中含有轉義字符例如   雙引號 ""  、大括號{}等

 

 

用stringByAddingPercentEncodingWithAllowedCharacters取代CFURLCreateStringByAddingPercentEscapes

1.網絡訪問請求：中文空格字符編碼/解碼

在 iOS 程序訪問 HTTP 資源時需要對 URL 進行 Encode，
比如像拼出來的 http://unmi.cc?p1=%+&sd f&p2=中文，
其中的中文、特殊符號&％和空格都必須進行轉譯才能正確訪問。
現在以“?!@#$^&%*+,:;='\”`<>()[]{}/\| "字符串為例子，
用stringByAddingPercentEncodingWithAllowedCharacters取代
CFURLCreateStringByAddingPercentEscapes

stringByAddingPercentEscapesUsingEncoding
(只對 `#%^{}[]|"<> 加空格共14個字符編碼，不包括“&?”等符號)，
iOS 9將淘汰，建議用stringByAddingPercentEncodingWithAllowedCharacters方法

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

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

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

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

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

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

2.網絡訪問請求：中文空格字符解碼
stringByRemovingPercentEncoding :
Xcode7可能會提示要將stringByAddingPercentEscapesUsingEncoding替換成此方法，要根據是否是解碼來區分

//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]; } 

舉例：

原字符串為：

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

CFURLCreateStringByAddingPercentEscapes方法為：

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

stringByAddingPercentEncodingWithAllowedCharacters方法為：

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

之后獲得的字符串為一致的。

問題實例：

我有一個有效的 url，我得到 不受支持的 url 錯誤。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} 

這是我想試圖如何初始化 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"]]; 

解決方法 :

URL 不能包含 ASCII 字符集中, 不是必須這樣的字符進行轉義的字符。
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); 

輸出:

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

作者：Kingsleeeey
鏈接：https://www.jianshu.com/p/dbb15f019127
來源：簡書
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