ERROR :- Undefined variable: raw While Transfering data between 2 pages [duplicate] - php

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"Notice: Undefined variable", "Notice: Undefined index", "Warning: Undefined array key", and "Notice: Undefined offset" using PHP
(29 answers)
Closed 4 years ago.
manageleavetype.php
Below code is my first file page I want to redirect this page form content into next file code edit.php. I want want to update the data from the next page.
<form action="" method="POST">
<tr>
<td><?php echo $type; ?></td>
<td><?php echo $des; ?></td>
<td><?php echo $time; ?></td>
<input type="hidden" name="hid" value="<?php echo $id; ?>">
<input type="hidden" name="htype" value="<?php echo $type; ?>">
<input type="hidden" name="hdes" value="<?php echo $des; ?>">
<td><input type="submit" name="del" value="DELETE">
<input type="submit" name="update" value="UPDATE">
</td></form></tr>
edit.php
This file except the code from above file and update the data that is received into MySQL database
<?php
require 'config.php';
if (isset($_POST['update'])) {
$eid = $_POST['hid'];
$que = "SELECT * FROM leavetype WHERE id='$eid'";
$exe = mysqli_query($conn, $que);
$raw = mysqli_fetch_array($exe);
}
if (isset($_POST['update'])) {
$nwtype = $_POST['ltype'];
$nwdes = $_POST['ldes'];
$que = "UPDATE leavetype SET leavetype='$nwtype',description='$nwdes' WHERE id='$eid'";
$exe = mysqli_query($conn, $que);
if ($exe) {
header("location:manageleavetype.php");
}
}
?>
<form method="POST">
<label>Leave Type</label>
<input type="text" name="ltype" value="<?php echo $raw['leavetype']; ?>">
<label>Description</label>
<input type="text" name="ldes" value="<?php echo $raw['description']; ?>">
<input type="submit" name="update" value="Update">
</form>
When I run this program following error is coming:
Notice: Undefined variable: raw in C:\wamp64\www\eLeaveSystem\edit.php on line 31

Answer 1

When you only define variables within an if block, you risk getting that error. In your case you define $raw only when update is posted:
if (isset($_POST['update'])) {
$eid = $_POST['hid'];
$que = "SELECT * FROM leavetype WHERE id='$eid'";
$exe = mysqli_query($conn, $que);
$raw = mysqli_fetch_array($exe);
}
But you access that variable unconditionally in the HTML generation part, like here:
<input type="text" name="ltype" value="<?php echo $raw['leavetype']; ?>">
So you have a few options:
Put the HTML generation part also in such an if block. Then you need to have another part in your code where you generate output for when no update is posted.
Define default values for those variables. For example $raw = ['leavetype' => 'something', 'description' => 'default description'].

Answer 2

You can simply solve this issue by using the isset() function to avoid undefined variable issue.
Try below:
<form method="POST">
<label>Leave Type</label>
<input type="text" name="ltype" value="<?php echo (isset($raw['leavetype']) ? $raw['leavetype'] : '');?>">
<label>Description</label>
<input type="text" name="ldes" value="<?php echo (isset($raw['description']) ? $raw['description'] : ''); ?>">
<input type="submit" name="update" value="Update">
</form>
I assume that the form block should appear for both Update and Insert.