java - 从 Java 调用 PHP 时的 Json 格式问题

Question

我在Java中调用mysql数据如下:

JSONObject jo = new JSONObject(result).getJSONArray("server_response").getJSONObject(0);
                if (jo.getString("success").equals("1"))
....

PHP:

   $dbhost = 'localhost';
   $dbuser = 'xxxx';
   $dbpass = 'xxxx';
      $conn =  mysqli_connect($dbhost, $dbuser, $dbpass,'test_db');

   if(! $conn ) {
          $response["success"] = 0;
        $response["message"] = "No connection.";
         header('Content-type: application/json');
   echo json_encode(array("server_response"=>$response));
        exit;
   }

   $sql = "Select * from employee ".
      "where 1=1 ";


  // array for JSON response
$response = array();    
   $retval = mysqli_query($conn,$sql );

   if(! $retval ) {

         $response["success"] = 0;
        $response["message"] = "User not found.";
   }
   else{       

        $response["success"] = 1;
        $response["message"] = "User has been found.";

        // echoing JSON response


   }
   header('Content-type: application/json');
   echo json_encode(array("server_response"=>$response));

在浏览器中获取结果为:

{"server_response":{"success":1,"message":"User has been found."}}

但在我看来，没有获得 Java 可以解释的正确 json 格式的值，因此代码无法进一步执行。

Answer 1

将 PHP 文件最后一行修改为:

echo json_encode(array("server_response"=>array($response)));