php通过JSON响应发布变量[重复]

Question

This question already has an answer here:

• “Notice: Undefined variable”, “Notice: Undefined index”, and “Notice: Undefined offset” using PHP 28 answers

Does anyone has an Idea how to get the POST value send to the php file through the same php response wrapped with JSON?

What I did so far was to send username and password to server side authentication. But it triggers an error of field missing. So I want to check the received data at the server side at the console.

  if(isset($_POST['u_name']) && isset($_POST['u_pass'])){

       $username = $_POST['u_name'];
       $pass = $_POST['u_pass'];


  } else {
       // required field is missing
       $response["success"] = 0;
       $response["message"] = "Required field(s) is missing" + $_POST['u_name'] + $_POST['u_name']; // <--- this is my line 48
       // echoing JSON response
       echo json_encode($response);

  }  

But I 'm getting an error like this

   <br />
   <b>Notice</b>:  Undefined index: u_name in <b>C:\xampp\htdocs\TestCordova\login_check.php</b> on line <b>48</b><br />
   <br />
   <b>Notice</b>:  Undefined index: u_name in <b>C:\xampp\htdocs\TestCordova\login_check.php</b> on line <b>48</b><br />
   {"success":0,"message":0}

</div>

Answer 1

Try this

<?php
if(isset($_POST['u_name']) && isset($_POST['u_pass'])){
   $username = $_POST['u_name'];
   $pass = $_POST['u_pass'];
} else {
   // required field is missing
   $response["success"] = 0;
   $string = "";
   if(!isset($_POST['u_name'])
    $string = "User name";
   if(!isset($_POST['u_pass'])
    $string.= " Password";

   $response["message"] = "Required field(s) is missing ".$string ; // <--- this is my line 48
   // echoing JSON response
   echo json_encode($response);

}  
?>