Question

我使用这种API登录我的应用程序。

NSString *url = [NSString stringWithFormat:@"%@//login",xyz];
    NSMutableURLRequest *request = [NSMutableURLRequest requestWithURL:[NSURL URLWithString:url]];


    NSError *error;
    NSData *jsonData = [NSJSONSerialization dataWithJSONObject:dict options:NSJSONWritingPrettyPrinted error:&error];
    if (!jsonData) {
        NSLog(@"Error creating JSON object: %@", [error localizedDescription]);
    }


    [request setValue:@"application/json;charset=utf-8" forHTTPHeaderField:@"Content-Type"];
    [request setValue:APIKEY forHTTPHeaderField:@"X_API_KEY"];

    [request setHTTPMethod:@"POST"];
    [request setHTTPBody:jsonData];

    [NSURLConnection sendAsynchronousRequest:request
     // the NSOperationQueue upon which the handler block will be dispatched:
                                       queue:[NSOperationQueue mainQueue]
                           completionHandler:^(NSURLResponse *response, NSData *data, NSError *error)
     {
         NSHTTPURLResponse *httpResponse = (NSHTTPURLResponse *)response;

         NSDictionary *responseDict = [NSJSONSerialization JSONObjectWithData: data options: 0 error: &error];  //I am using sbjson to parse

         if(httpResponse.statusCode == 200)
         {

             //Show message to user success
         }
         else  if(httpResponse.statusCode == 401)
         {
            //Show message to user -fail
         }
         else if(httpResponse.statusCode == 500)
         {
             //Show message to user- server error
         }
     }];

当我使用正确的用户名和密码时，我得到httpResponse和状态代码为200.但如果使用错误的用户名和密码，我没有得到401。

那么如何处理这种情况

此致兰吉特

Answer 1

您正在将NSURLResponse投射到NSHTTPURLResponse，但这不会自动公开statusCode属性中您期望的实际值。

相反，您应该检查是否已填充error属性并检查code属性。

if(error)
{
    NSLog(@"error: %@", error.code);

    if(error.code == 401)
    {
        //handle 401 errors
    }
    else if(error.code == 500)
    {
        //handle 500 errors
    }
}
else
{
    //successful request
}

使用sendAsynchronousRequest处理401状态代码：iOS中的请求API

1 个答案: