Undefined variable $id everytime I use $_POST method

Question

I'm using post method to get the id but everytime tells me Undefined variable $id here is my code:

<?php    
    require 'DBConnection.php';
    $code='';
    if(isset($_POST["ID"])){
        if($_POST(["ID"]) != ''){
            $id = $_POST['ID'];
            $get_c = $pdo->prepare("SELECT * FROM all_menu WHERE `ID` = '".$id."'");
            $get_c->execute(); 
            while ($row = $get_c->fetch()) {
                $code .= $row['item_code'];
            }
            echo $code;
        }
    }

Edit: I'm using this code to select from dropdown options to get the ID of the value in dropdown for example I have three options in my select dropdown let's say the first one is "Pepsi" when I select "Pepsi" I need to put its ID in an input I'm using PHP and Ajax in this code.

Here is my Ajax code:

<script>
  $(document).ready(function(){
    $('#itemname').change(function(){
      var code = $(this).val();  
       $.ajax({
          type: 'POST',
          url: 'pages/GetCode.php',
          data:{code:code},
          success: function(data){
            document.getElementById("itemcode").value = data;
          },
          error: function (jqXHR, textStatus, errorThrown){ 
            alert(errorThrown);
          }
       });
    });
  });
</script>

Try to echo $id before query and check what value $id is holding . — Sho, Oct 17 '21 at 06:03
$id before query holding that: Uncaught Error: Array callback must have exactly two elements — Hazem El-behairy, Oct 17 '21 at 06:15
Why do you use a prepared statement, but string concatenation for the query? It is widely open for SQL injection — Nico Haase, Oct 17 '21 at 06:46
Also, does that error message really read "Undefined variable $id"? — Nico Haase, Oct 17 '21 at 06:47

score 1 · Answer 1 · answered Oct 17 '21 at 06:33

If you are calling this php page from the javascript AJAX, then I can see the only parameter is "code", so how can you expect "ID" in the POST?

Either change the javascript, or the PHP file.

The below javascript will send "code" as POST parameter, data:{"code":code},

<script>
  $(document).ready(function(){
    $('#itemname').change(function(){
      var code = $(this).val();  
       $.ajax({
          type: 'POST',
          url: 'pages/GetCode.php',
          data:{"code":code},
          success: function(data){
            document.getElementById("itemcode").value = data;
          },
          error: function (jqXHR, textStatus, errorThrown){ 
            alert(errorThrown);
          }
       });
    });
  });
</script>

Use "code" instead of "ID" as POST parameter in PHP

<?php    
    require 'DBConnection.php';
    $code='';
    if(isset($_POST["code"])){
       if($_POST(["code"]) != ''){
            $id = $_POST['code'];

Side Note: Do not use user input (here "code") in the SQL query, that way your sql query will be open to injection.

score 1 · Answer 2 · answered Oct 17 '21 at 06:44

1

You are sending param code as ajax data and you are fetching param id Change $_POST['id'] to $_POST['code']

$id = $_POST['code'];

answered Oct 17 '21 at 06:44

Sho

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Undefined variable $id everytime I use $_POST method

2 Answers2