undefined index error in my php script for deleting user

Question

im trying to make a delete user page but I cannot figure out why I am getting this error.

Undefined index: bil in C:\wamp\www\FORM\deleteadmin.php on line 4

Undefined index: adminID in C:\wamp\www\FORM\deleteadmin.php on line 5

which is line 4 and 5 looks like this,

$bil = $_POST['bil'];
$adminID = $_POST['adminID'];

the full code are new.php which links to the deleteadmin.php that contains the error.

new.php

echo "<form name='update $bil' method=post action=newuser.php>" ?>
    <tr>
        <td><center><?php echo $bil; ?></center></td>
        <td><?php echo $row['name']; ?></td>
        <td><?php echo $row['tel']; ?></td>
        <td><?php echo $row['email']; ?></td>
        <td><?php echo $row['username']; ?></td>
        <td><?php echo $row['password']; ?></td>
        <td><a href = "deleteadmin.php"><img src="img/deleteicon.png" align="center"></a></td>
        <input type=hidden value={$row['adminID']} name=adminID>
        <input type=hidden value=$bil name=bil>
    </tr>
<?php 
    echo "</form>";
}
?>

deleteadmin.php

<?php
include ('config.php');

$bil = $_POST['bil'];
$adminID = $_POST['adminID'];

$result = mysql_query("DELETE FROM `admin` WHERE `adminID`= '$adminID'") or die (mysql_error());

if($result)
{
?>
<script language="javascript">
alert("User No : <?php echo "$bil"; ?> DELETED");
location.href="new.php";
</script>

<?php
}
else
{
?>

<script language="javascript">
alert("User No : <?php echo "$bil"; ?> NOT DELETED");
location.href="new.php";
</script>
<?php
}
?>

What do I need to do to fix them? I've tried using ISSET but it doesnt work. I don't understand why this is happening, and I'd love to know how to make it go away.

1. You should always surround values in HTML with quotes as a matter of good practice and it sometimes causing problems. 2. How are you submitting this form? I don't see a submit button. — Jeff
– Jeff, Commented Apr 1, 2014 at 6:15
action=newuser.php are you submitting to the right script? — esqew
– esqew, Commented Apr 1, 2014 at 6:16
@jeff what I want to make is when I clicked the 'X' button, the user will be deleted instantly — Syaa
– Syaa, Commented Apr 1, 2014 at 6:59
Just have the X button fire an AJAX POST call to deleteadmin.php with data: { "bil" : "1", "adminID", "1" }. jQuery.post() makes this easy. — Jeff
– Jeff, Commented Apr 1, 2014 at 7:02

Nikko Madrelijos · Accepted Answer · 2014-04-01 06:51:41Z

2

I see 2 problems here

First:

<a href = "deleteadmin.php"><img src="img/deleteicon.png" align="center"></a>

By clicking this you are only accessing the deleteadmin.php file. You are not passing any arguments.

Second:

$bil = $_POST['bil'];
$adminID = $_POST['adminID'];

You are not posting any data so the $_POST variable will be empty.

Solutions

Add query string in the url ?bil=".$bil."&adminID=".$adminID."

<a href = "deleteadmin.php?bil=".$bil."&adminID=".$adminID.""><img src="img/deleteicon.png" align="center"></a>

Pass get Values then process in the deleteadmin.php Instead of using $_POST use $_GET.

$bil = $_GET['bil'];
$adminID = $_GET['adminID'];

$_GET[] is used to get the values which are encoded in the query string in the URL.

Then you don't need to have a form for you to delete a user. Just access the deleteadmin.php file passing $_GET values in the url.

edited Apr 1, 2014 at 6:51

answered Apr 1, 2014 at 6:24

Nikko Madrelijos

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3 Comments

Nikko Madrelijos Over a year ago

Also in the query "DELETE FROM admin WHERE adminID= '$adminID'" This is not safe. To make it more secure always clean user input. In this case md5 encryption is more than enough "DELETE FROM admin WHERE md5(adminID)= '".md5($adminID)."'" This helps clean user input. It encrypts the user input so that any unwanted characters will not cause any error or sql injection.

Syaa Over a year ago

i did what you suggest, but it still shows error on line 5.

Nikko Madrelijos Over a year ago

Check the variables in the link.. They should have exactly the same variable. if you put adminID in the url, in the php you should use $_GET['adminID'] too.

user3462511 · Accepted Answer · 2014-04-01 06:26:13Z

2

this is very easy to fix.

try to do not use hidden and replace your href like this in your new.php:

    <td><a href = "deleteadmin.php?adminID={$adminID}&bil={$bil}"><img src="img/deleteicon.png" align="center"></a></td>

actually you can replace $_POST to become $_REQUEST because $_REQUEST represent $_POST and $_GET

hope this can help you to fix your problem.

answered Apr 1, 2014 at 6:26

user3462511

Comments

Shankar Narayana Damodaran · Accepted Answer · 2014-04-01 07:00:31Z

I don't think what you are doing is right.. Because, when you click the <a> link this will going to give you undefined index errors. So you should pass the parameters as a querystring to your <a> .. that is because the <form> action points to newuser.php

You need to append the parameters , adminID and bil to this link like this..

<td><a href = "deleteadmin.php?adminID=<?php echo $row['adminID']; ?>&bil=<?php echo $bil; ?>"><img src="img/deleteicon.png" align="center"></a></td>

and on your deleteadmin.php , you need to change from

$bil = $_POST['bil'];
$adminID = $_POST['adminID'];

to

$bil = $_GET['bil'];
$adminID = $_GET['adminID'];

This(mysql_*) extension is deprecated as of PHP 5.5.0, and will be removed in the future. Instead, the MySQLi or PDO_MySQL extension should be used. Switching to PreparedStatements is even more better to ward off SQL Injection attacks !

Remya Murali · Accepted Answer · 2014-04-01 06:19:52Z

0

change code in new.php to:

echo "<form name='update $bil' method=post action=newuser.php>
    <tr>
        <td><center><?php echo $bil; ?></center></td>
        <td><?php echo $row['name']; ?></td>
        <td><?php echo $row['tel']; ?></td>
        <td><?php echo $row['email']; ?></td>
        <td><?php echo $row['username']; ?></td>
        <td><?php echo $row['password']; ?></td>
        <td><a href = 'deleteadmin.php'><img src='img/deleteicon.png' align='center'></a></td>
        <input type='hidden' value=$row['adminID'] name='adminID'>
        <input type=hidden value='$bil' name='bil'>
    </tr>
</form>";

try this.

answered Apr 1, 2014 at 6:19

Remya Murali

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