php - JsonObjectRequest 不发送或接收 PHP 参数

Question

当我阅读响应时，我总是从 PHP 收到错误消息：查询中缺少参数。我知道 PHP 和查询 100% 很好，因为它可以工作。只会以这种方式发生。

Volley.class

JSONObject params = new JSONObject();
params.put("consulta", "UPDATE Usuarios SET Nombre='AAA' WHERE IdUsuario=1");
//All right

RequestFuture<JSONObject> future = RequestFuture.newFuture();
JsonObjectRequest request = new JsonObjectRequest(Request.Method.POST, URL", params, future, future); //request contains params with de sql sentence. Its well
VolleyS.getInstance(context).mRequestQueue.add(request);

JSONObject response = future.get(10, TimeUnit.SECONDS); //response contains the wrong error and the database isn't changed
if (response.getInt("success") == 0) correctoExterna = false;
else correctoExterna = true;

更新.php

if (isset($_REQUEST['consulta'])){

$vConsulta= $_REQUEST['consulta'];
$result = mysql_query($vConsulta);

// mysqli_errno($db) != 0
if ($result)
    {
       $response["success"] = 1;
       $response["message"] = "OK.";
       echo json_encode($response);
       myqli_commit($db);
       mysqli_close($db);

    }
    else{
        $response["success"] = 0;
        $response["message"] = "ERROR :".mysqli_error($db);
        echo json_encode($response);    
        mysqli_rollback($db);
        mysqli_close($db);

    }
 }
 else {
     $response["success"] = 0;
     $response["message"] = "ERROR, missing fields to insert";
     echo json_encode($response);
 mysqli_close($db);
 }