Why I get "Notice: Undefined index" error in PHP

Question

I got an error in this section, I don't have any ideas to fix this.

Notice: Trying to access array offset on value of type null in C:\xampp\htdocs\web_resto\pagePelanggan.php on line 16

Notice: Undefined index: status_order in C:\xampp\htdocs\web_resto\pagePelanggan.php on line 26

Code:

<?php
include "config/controller.php";
$id = new Resto();
session_start();
$auth     = $id->AuthUser($_SESSION['username']);
$auth2    = $id->AuthPelanggan($_SESSION['username']);
$response = $id->sessionCheck();
if ($response == "false") {
    header("Location:index.php");
}
$no_meja = $auth2['no_meja'];
$sql2    = "SELECT kd_order FROM tb_order WHERE no_meja='$no_meja'";
$exe2    = mysqli_query($con, $sql2);
$num2    = mysqli_num_rows($exe2);
$dta2    = mysqli_fetch_assoc($exe2);
$data_kd = $dta2['order_kd'];
$sql3     = "SELECT status_detail FROM tb_detail_order_temporary WHERE order_kd='$data_kd'";
$exe3     = mysqli_query($con, $sql3);
$num3     = mysqli_num_rows($exe3);
$dta3     = mysqli_fetch_assoc($exe3);
$data_kd2 = $dta3['status_detail'];
$sql4     = "SELECT status_order FROM tb_order WHERE kd_order='$data_kd'";
$exe4     = mysqli_query($con, $sql4);
$num4     = mysqli_num_rows($exe4);
$dta4     = mysqli_fetch_assoc($exe4);
$data_kd3 = $dta4['status_order'];
if (isset($_GET['delete'])) {
    if ($data_kd3 == "belum_beli") {
?>

Please understand what the message actually says. This is a _notice_ from a logical point of view, not an error. You try to access an element of an array which does not exist. That is all. — arkascha, Aug 07 '21 at 09:41
Don't post [images of code](https://meta.stackoverflow.com/questions/285551/why-not-upload-images-of-code-errors-when-asking-a-question)! — brombeer, Aug 07 '21 at 09:59
Does this answer your question? ["Notice: Undefined variable", "Notice: Undefined index", and "Notice: Undefined offset" using PHP](https://stackoverflow.com/questions/4261133/notice-undefined-variable-notice-undefined-index-and-notice-undefined) — Tim Yao, Aug 07 '21 at 12:14

KIKO Software · Answer 1 · 2021-08-07T10:44:09.390

There are many things wrong with your code, but the issue, at this moment, is that your query is:

SELECT kd_order FROM tb_order WHERE no_meja='$no_meja'

but in your PHP code you ask for:

$data_kd = $dta2['order_kd'];

You have kd_order in your query, but you use order_kd as the index. That will never work.

Other (major) problems are:

It is strongly advised not to put PHP variables directly into query strings.
You don't check whether your queries return anything. This is probably the cause of the first notice. It is not unlikely for a query to return NULL.

If you enjoy PHP, have a look at this website: PHP: The Right Way

Why I get "Notice: Undefined index" error in PHP

1 Answers1