php post variables through JSON response

Question

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}

... Well, you are quiet literally checking if the values are there, and if not, then throwing a error, why are you surprised that you get a error? — Epodax, May 02 '16 at 10:33
`else` is executing because `$_POST['u_name']` is not there. So how can you except the value — urfusion, May 02 '16 at 10:34
you cannot use $_POST['u_name'] in else because if it goes to else that mean $_POST['u_name'] isn't set — Madhawa Priyashantha, May 02 '16 at 10:34
@epodax I'm not suprised. I just want to check what are those values I'm sending. Hope you got what I really want from this? — Thush-Fdo, May 02 '16 at 10:35
Just use "Required field(s) is missing" OR "Username and Password fields missing". You can not use same variable in else condition of isset. — RJParikh, May 02 '16 at 10:36
@riggsfolly It just check whether the $_POST() variable is set for a value....? Am I correct? — Thush-Fdo, May 02 '16 at 10:40
Yes, so if `isset($_POST['u_name'])` returns false you cannot then use `$_POST['u_name']` in an error message in the ELSE! Apart from the obvious thing that you already know it does not have a value, **it does not actually exist** — RiggsFolly, May 02 '16 at 10:42
So as @RuchishParikh tried to say, just output a generalised error message saying `"Username and Password fields missing"` — RiggsFolly, May 02 '16 at 10:45
**Side Note:** `isset()` can take more than one parameter so your IF can be coded as `if ( isset($_POST['u_name'], $_POST['u_pass']) ) {` — RiggsFolly, May 02 '16 at 10:46

score 0 · Answer 1 · answered May 02 '16 at 13:14

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);

}  
?>

php post variables through JSON response

1 Answers1