Question

我必须解析JSON API的返回，我已经将URL作为一个字符串，然后我将它传递给NSURL，但是在NSLog中它显示为nil。这是我的代码，

NSString *urlstring=[NSString stringWithFormat:@"http://api.v3.factual.com/t/places?q=%@&geo={\"$circle\":{\"$center\":[19.9909631,73.8034808],\"$meters\":40000}}&limit=20&KEY=123456",[self.strurl lowercaseString]];
NSURL *url=[[NSURL alloc]initWithString:urlstring ];
NSURLRequest *req = [NSURLRequest requestWithURL:url];

提前致谢。

Answer 1

您必须对您的网址进行编码，因为它包含特殊字符，请尝试使用此类

NSString *paramString=[NSString stringWithFormat:@"/t/places?q=%@&geo={\"$circle\":{\"$center\":[19.9909631,73.8034808],\"$meters\":40000}}&limit=20&KEY=123456",[self.strurl lowercaseString]];

NSString *encodedSearchString = [paramString stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding];

NSString *urlString = [NSString stringWithFormat:@"http://api.v3.factual.com%@", encodedSearchString];
NSURL *url = [NSURL URLWithString:urlString];
NSLog(@"urlstring is %@",url)

希望这有帮助

Answer 2

试试，测试和工作

-(void) sample
{
   // NSString *strurl = @"hello";
    NSString *urlstring=[NSString stringWithFormat:@"http://api.v3.factual.com/t/places?q=%@&geo={\"$circle\":{\"$center\":[19.9909631,73.8034808],\"$meters\":40000}}&limit=20&KEY=123456",[strurl lowercaseString]];
    NSURL *url=[NSURL URLWithString:[urlstring stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding]];
    NSURLRequest *req = [NSURLRequest requestWithURL:url];
    NSLog(@"%@",req);
}

在将NSString设置为NSURL时，NSURL返回nil值

2 个答案: