为 iphone 解析 json 时获取空值？

Question

我正在尝试在这里解析 JSON 并执行一些操作。我在字符串中得到响应，如下所示，为什么 json 返回 null，

    NSError *error = [[NSError alloc] init];
    NSHTTPURLResponse *response = nil;
    NSData *urlData=[NSURLConnection sendSynchronousRequest:request returningResponse:&response error:&error];

    NSLog(@"Response code: %d", [response statusCode]);
    if ([response statusCode] >=200 && [response statusCode] <300)
    {
        NSString *responseData = [[NSString alloc]initWithData:urlData encoding:NSUTF8StringEncoding];
        NSLog(@"Response ==> %@", responseData);

我正在这里收到回复...

响应 ==> {"success":1}{"tag":"login","success":1,"error":0}

如果响应可以以字符串形式出现，为什么它不会出现在下面的代码中？我们可以在下面的代码中获取成功函数/变量或将字符串传递给jsonData...

 NSDictionary *jsonData = [NSJSONSerialization JSONObjectWithData:urlData options:NSJSONReadingMutableLeaves error:nil];
                          NSLog(@"json data is %@",jsonData);
            NSInteger success = [(NSNumber *) [jsonData objectForKey:@"success"] integerValue];
            NSLog(@"%d",success);

        if(success == 1)
        {
            NSLog(@"Login SUCCESS");
            [self alertStatus:@"Logged in Successfully." :@"Login Success!"];
                   ColorPickerViewController *cpvc =[[ColorPickerViewController alloc] init];
                   [self.navigationController pushViewController:cpvc animated:YES];

        } else {

            NSString *error_msg = (NSString *) [jsonData objectForKey:@"error_message"];
            [self alertStatus:error_msg :@"Login Failed!"];
        }

每次我运行它都会执行 else 块...

当我尝试解析 Json 时，它正在重新调整我的空值

2013-12-18 07:52:09.193 ColorPicker[15867:c07] json 数据为 (null) 2013-12-18 07:52:09.193 ColorPicker[15867:c07] 0

我在这里发送来自 json 的响应

 // Get tag
   $tag = $_POST['tag'];

   // Include Database handler
   require_once 'include/DB_Functions.php';
   $db = new DB_Functions();
   // response Array
   $response = array("tag" => $tag, "success" => 0, "error" => 0);

   // check for tag type
   if ($tag == 'login') {
       // Request type is check Login
       $email = $_POST['email'];
       $password = $_POST['password'];

       // check for user
       $user = $db->getUserByEmailAndPassword($email, $password);
       if ($user != false) {
           // user found

           // echo json with success = 1
           $response["success"] = 1;
           $response["user"]["fname"] = $user["firstname"];
           $response["user"]["lname"] = $user["lastname"];
           $response["user"]["email"] = $user["email"];
   $response["user"]["uname"] = $user["username"];
           $response["user"]["uid"] = $user["unique_id"];
           $response["user"]["created_at"] = $user["created_at"];

           echo json_encode($response);
       } else {
           // user not found
           // echo json with error = 1
           $response["error"] = 1;
           $response["error_msg"] = "Incorrect email or password!";
           echo json_encode($response);
       }
   }

我有几乎相同的 web 服务用于注册标签注册，并且使用相同的代码可以正常工作：O

Answer 1

此回复：

响应 ==> {"success":1}{"tag":"login","success":1,"error":0}

似乎有两个有效的 json 字符串：

• {“成功”：1} 和
• {"tag":"login","success":1,"error":0}

所以，您的 php 代码可能在两个地方写出 json。调试脚本， 例如，通过为成功设置不同的值（两个 json 字符串中的公共键）， 应该有助于确定它。

Answer 2

如果您的 jsonData 为零。尝试这样解析

如果是数组，你的响应应该是这样的：

Response : [{"success":1},{"tag":"login","success":1,"error":0}]

如果是字典，那么响应应该是

Response : {"tag":"login","success":1,"error":0}

然后替换这个

    NSDictionary * jsonData = [NSJSONSerialization JSONObjectWithData:urlData options:kNilOptions error:&error];


    NSInteger success = [(NSNumber *) [jsonData objectForKey:@"success"] integerValue];

有

    NSInteger success = [[jsonData objectForKey:@"success"] integerValue];

或

    NSInteger success = [[jsonData valueForKey:@"success"] integerValue];

或

    NSInteger success = [jsonData[@"success"] integerValue];