delete mysql_array_fetch - mysql

Question

I have the following code and the plan is to display the data from the database but allow for the administrator of the site to delete a row if a job is no longer available. I have put "Delete" where I would like a link to delete the row.I have tried using <a href='delete1.php?del=$row[JobID]'>Delete</a> but that just throws an error up on the page.
<?php
include_once('db.php');
$result = mysqli_query($con,"SELECT * FROM Job ORDER BY JobID");
echo "<table border='1'>
<tr>
<th>Job ID</th>
<th>Job Title</th>
<th>Job Description</th>
<th>Industry</th>
<th>Job Type</th>
<th>Salary</th>
<th>County</th>
<th>Town</th>
<th>Delete</th>
</tr>";
while($row = mysqli_fetch_array($result))
{
echo "<tr>";
echo "<td>" . $row['JobID'] . "</td>";
echo "<td>" . $row['JobTitle'] . "</td>";
echo "<td>" . $row['JobDescription'] . "</td>";
echo "<td>" . $row['Industry'] . "</td>";
echo "<td>" . $row['JobType'] . "</td>";
echo "<td>" . $row['Salary'] . "</td>";
echo "<td>" . $row['County'] . "</td>";
echo "<td>" . $row['Town'] . "</td>";
echo "<td>" . "Delete" . "</td>";
echo "</tr>";
}
echo "</table>";
mysqli_close($con);
?>

Answer 1

You should put this in the the listing code:
echo "<a href='delete1.php?del={$row['JobID']}'>Delete</a>";
(documentation)
Then, in your delete1.php you should have something like:
$jobid = intval($_GET['JobID']);
if ($jobid > 0) {
mysqli_query($con, "DELETE FROM Job WHERE JobID=$jobid LIMIT 1");
}
(note: this is untested and can be quite insecure; it only shows the concept on how to do this)