php代码2响应不打印

Question

我尝试使用post请求从我的表barrel_user中获取player id的值，我唯一的问题是两个响应不起作用，但是如果我只给一个，它就能工作了。

        $customer_query="SELECT player_id FROM barrel_user where     phone_number='$phone_number'";
$fetchresult=mysqli_query($conn,$customer_query);

  if(mysqli_num_rows($fetchresult)>0)
{
  while($rows = mysqli_fetch_assoc($fetchresult)){

//  $response=$rows;


  if ($fetchresult) {
        $response["success"] = 1;
        //$response["player_id"]= $player_id;

            $player_id= $rows['player_id']; 
              $response=$rows;
    } else {
        $response["success"] = 0;
    }

  // $response["success"] = 'User Exists';
  }
}

上面是我的post请求的php代码。

我没有得到这个值$response“成功”= 1；

但是我得到了player_id的价值

注意：-代码工作正常，如果我删除$response=$rows，我将得到消息success=1

Answer 1

尝尝这个

$response=array();
$customer_query="SELECT player_id FROM barrel_user where     phone_number='$phone_number'";
$fetchresult=mysqli_query($conn,$customer_query);

  if(mysqli_num_rows($fetchresult)>0)
{
  while($rows = mysqli_fetch_assoc($fetchresult)){

  if ($fetchresult) {
        $response["success"] = 1;
        //$response["player_id"]= $player_id;

            $player_id= $rows['player_id']; 
              $response[]=$player_id;
    } else {
        $response["success"] = 0;
    }

  }
}