Question

我在服务器上发出请求，在URL中有空格

http://xxxxxxxx.com/api/api.php?func=showAllwedsdwewsdsd&params [] =％沙特＆20Arab放大器; PARAMS [] =所有

我收到了错误的错误网址，

我用过

   downloadURL = [downloadURL stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding];

但是由于这个原因我得到了奇怪的网址

 http://sdsdsdsdsdsd.com/api/api.php?func=showAllsdsd&params5262721536=Saudi              0X0P+0rabia&params5 8288=All

我也在使用

  downloadURL= [downloadURL stringByReplacingOccurrencesOfString:@" " withString:@"%20"];

但又是一个奇怪的网址

     http://xxxxxxxxx.com/api/api.php?func=showsdsdsd&params[]=Saudi 0X1.240DC0D824P-807rabia&params[]=All

请帮助我如何解决此问题

编辑粘贴大部分代码

   PropertyXMLDownloader *download=[[PropertyXMLDownloader alloc]init];
  [download setDelegate:self];
   firstAppDelegate *del=(firstAppDelegate *)[[UIApplication sharedApplication]delegate];
   NSLog(@"Country is %@",del.country);
   NSLog(@"State is %@",del.state);
//  del.country=[del.country stringByReplacingOccurrencesOfString:@" " withString:@"%20"];

   NSString *downloadURL=[NSString stringWithFormat:@"http://xxxxxxxx.com/api/api.php?func=showAll&params[]=Saudi Arabia&params[]=%@",@"all"];
      // downloadURL= [downloadURL stringByReplacingOccurrencesOfString:@" " withString:@"%20"];
       //downloadURL = [downloadURL stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding];

   NSCharacterSet *doNotWant = [NSCharacterSet characterSetWithCharactersInString:@" "];
   downloadURL = [[downloadURL componentsSeparatedByCharactersInSet: doNotWant]    componentsJoinedByString:@"%20"];


    NSLog(downloadURL); 
    [download startDownloading:downloadURL];

Answer 1

试试这个。

NSCharacterSet *doNotWant = [NSCharacterSet characterSetWithCharactersInString:@" "];
downloadURL = [[downloadURL componentsSeparatedByCharactersInSet: doNotWant]   componentsJoinedByString:@"%20"];

Answer 2

也许％20被视为数据参数，例如％@或％g。尝试使用

定义NSString

NSString *urlString = [NSString stringWithFormat:@"http://xxxxxxxx.com/api/api.php?func=showAllwedsdwewsdsd&params[]=Saudi%20Arab&params[]=all"];

你会看到一个警告。通过在其前面添加另一个符号来“转义”百分号：

NSString *urlString = [NSString stringWithFormat:@"http://xxxxxxxx.com/api/api.php?func=showAllwedsdwewsdsd&params[]=Saudi**%**%20Arab&params[]=all"];

并且警告消失了。

Answer 3

你的问题是：

NSLog(downloadURL);

尝试将其替换为：

NSLog(@"%@", downloadURL);

一切都会奏效。

您可以使用stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding并忘记所有解决方法。

（解释：由于downloadURL包含％符号，因此它不能很好地与NSLog一起使用，后者期望将格式字符串作为其第一个参数，其中％符号标识要替换的占位符。

创建NSURL时的URL错误

3 个答案: