Question

我有以下代码;

// check phone exist or not
   $query = "SELECT * FROM user WHERE phone_number=".$phone;
   $result = mysqli_query($conn,$query);
   $count = mysqli_num_rows($result);
   if($count!=0){
    $error = true;
    $phoneError = "Provided Phone Number($phone) is already in use.";
   }

如何将查询结果作为json feed发送给我们同一个项目的同事？

Answer 1

您需要创建一个数组，您可以在其中添加fail/success等状态以及基于此的相应消息。

下面给出了一个示例代码： -

$final_message = []; 
if($count > 0){ 
  $final_message['status'] = 'fail'; 
  $final_message['message'] = "Provided Phone Number($phone) is already in use."; 
}else{ 
  $final_message['status'] = 'success'; 
} 
echo json_encode($final_message);

将mysql数据作为json feed发送给另一个开发人员

1 个答案: