undefined index of php $id get value of the variable

Question

<?php
include 'connect.php';
$conn = mysqli_connect($host, $user, $pass, $db);

if (!$conn) {
    die("KO" . mysqli_connect_error());
}

$id = $_GET['id'];

$sql = "select * from tblabc where id = '$id'";

if (mysqli_num_rows($sql)==0){
    echo 'ok';
} else {
    $data = mysqli_fetch_assoc($sql);
}
?>

this is my code to editing the table record. but when its executed occurs error "notice: Undefined index: id". how to resolved this issued. thx

Are you sure you are passing the `id` value to this page via `$_GET` ? What does `var_dump($_GET['id'])` give you ? — Maximus2012, Oct 28 '16 at 16:56
your code is vulnerable to SQL-injection just imagine (don't try) someone asking for the id `';DROP TABLE tblabc -- ` — Franz Gleichmann, Oct 28 '16 at 17:05
@FranzGleichmann: thx for ur advise, for study only not for real word usage — , Oct 28 '16 at 17:20
@mogleng the fact that it's not for study is no excuse to write insecure code - because that way, you *learn* to write insecure code. you should instead do it the right way from the very beginning. — Franz Gleichmann, Oct 28 '16 at 17:22

score 0 · Answer 1 · answered Oct 28 '16 at 17:02

0

<?php
include 'connect.php';
$conn = mysqli_connect($host, $user, $pass, $db);

if (!$conn) {
    die("KO" . mysqli_connect_error());
}
if(isset($_GET))
{
$id = $_GET['id'];

$sql = "select * from tblabc where id = '$id'";

if (mysqli_num_rows($sql)==0){
    echo 'ok';
} else {
    $data = mysqli_fetch_assoc($sql);
}
}
else
{
 echo 'nothing received';
}
?>

Use the above code to check the validity of the get variable before using it.This way you'll be able to know where the error is coming from.

answered Oct 28 '16 at 17:02

Black Mamba

13,632
6
82
105

ur code response : Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result, string given in C:\xampp\htdocs\web\edit.php on line 30 ok – Oct 28 '16 at 17:17
please tell me the url of the page you are running this script in – Black Mamba Oct 28 '16 at 18:03
http://localhost/web/edit.php – Oct 29 '16 at 17:23
solved. thx for the assistant – Oct 30 '16 at 05:25

score 0 · Answer 2 · answered Oct 28 '16 at 17:03

0

you should try with this

$id  = '';
if(isset($_GET['id']) && $_GET['id'] != '')
{
    $id = $_GET['id']; 
    $sql = "select * from tblabc where id = '$id'";

    if (mysqli_num_rows($sql)==0){
        echo 'ok';
    } else {
        $data = mysqli_fetch_assoc($sql);
    }
}
else
{
echo 'Id Not Defined';
}

answered Oct 28 '16 at 17:03

Rohit

377
3
15

ur code respone id not defined. – Oct 28 '16 at 17:18
yes because of your $_GET['id'] is not set.. please check in URL or other source where form id come to this page – Rohit Oct 28 '16 at 17:20
and how to check it? – Oct 28 '16 at 17:27
solved. thx for the assistant – Oct 30 '16 at 05:25

undefined index of php $id get value of the variable

2 Answers2