How to solve Undefined index

Question

I have a form like this, if I choose the Admin level, then the combo box department is disabled, then when choosing the Staff level, then the combo box department will be active and display the department's options, this method works, but it encountered an error during the save process by selecting the Admin level, and the form department is not active,

This my code

    <div class="form-group">
            <label class="control-label col-sm-4" for="level">Level</label>
            <div class="col-sm-4"> 
              <select name="level" id="level" class="form-control">
              <option value="null">-- Level --</option>
              <option value="admin">Admin</option>
              <option value="staff">Staff</option>
              <option value="client">Client</option>
              </select>
            </div>
          </div>
          <div class="form-group">
            <label class="control-label col-sm-4" for="department">Department</label>
            <div class="col-sm-4"> 
              <select name="department" id="department" class="form-control">

              </select>
            </div>
          </div>
 $("#level").change(function(){

        // variabel dari nilai combo box provinsi
        var id = $("#level").val();


        // mengirim dan mengambil data
        $.ajax({
            type: "POST",
            dataType: "html",
            url: "search_level.php",
            data: "level="+id,
            success: function(msg){

                // jika tidak ada data
                if(msg == ''){
                  var x=document.getElementById("department")
                  x.value=""
                  x.disabled=true
                }               
                // jika dapat mengambil data,, tampilkan di combo box kota
                else{
                  var x=document.getElementById("department")
                  x.disabled=false
                    $("#department").html(msg);                                                     
                }
            }
        });    
    });

code save proses

   <?php

include "../config/koneksi.php";

$name = $_POST['name'];
$user = $_POST['username'];
$password = md5($_POST['password']);
$level = $_POST['level'];
$deparment = $_POST['department'];

$status = '1';
$tgl_dibuat = date("Y-m-d h:i:s");

$query = mysqli_query($con, "INSERT INTO user (nama,user,password,level,department,tgl_dibuat,status) VALUES ('$name','$user',
'$password','$level','$deparment','$tgl_dibuat','$status')");

if ($query) {
    ?>
    <script language="JavaScript">
    alert('User Saved');

    </script>
    <?php
} else {
    ?>
    <script language="JavaScript">
    alert('Failed');
    document.location='user';
    </script>
    <?php
} 
?>

code search

<?php
    include "../config/koneksi.php";

    $id = $_POST['level'];

    $query = mysqli_query($con, "SELECT * FROM department WHERE level='$id'");
    while($data_prov=mysqli_fetch_array($query)){   
    ?>
        <option value="<?php echo $data_prov["department"] ?>"><?php echo $data_prov["department"] ?></option><br>

    <?php
    }
    ?>

Answer 1

If the field is disabled you won't send $_POST['department']; so:

$deparment = $_POST['department'];

fails to load 'department' index of your post request. You could do:

$department = isset($_POST['department']) ? $_POST['department'] : "Default value";

Please notice that your code is quite unsafe since you don't do any check on values and you don't escape strings, someone could easily inject some SQL inside the form. See mysqli_real_escape_string() .

I would at least escape your strings:

$name = mysqli_real_escape_string($_POST['name']);
$user = mysqli_real_escape_string($_POST['username']);
$password = md5($_POST['password']);
$level = mysqli_real_escape_string($_POST['level']);
$deparment = isset($_POST['department']) ? mysqli_real_escape_string($_POST['department']) : "Default value";

Also, as @alexis observe, it would be even better to use Prepared statements .

Answer 2

Basically this error tells you that in file "add_user.php" on line 9 you are trying to print index which actually doesn't exist. That line of code is given below...

<option value="<?php echo $data_prov["department"] ?>"><?php echo $data_prov["department"] ?></option><br >

So, in order to solve this problem, you have to find out what are columns name in your database table department . Just type or copy paste exactly same these column name instead of department into given below code line...

<option value="<?php echo $data_prov["column_name"] ?>"><?php echo $data_prov["column_name"] ?></option><br`>

Then this error will be resolved.