lukecolli98 lukecolli98 - 11 months ago 30
HTML Question

How to use PHP Variables in an SQL statement

Before anybody says, I will protect myself against SQL injections, right after I fix this error. I am making an app where news reports are submitted to the database. This page is what removes a report from the database.

What I have tried:
Every possible way of adding brackets to the SQL and speech marks. My ICT teacher and I have looked at this for nearly 2 hours and cannot find a fix. I have also searched Google and Stack Overflow but I cannot find an answer.

Ok, so the correct report_id displays when I echo it. When I put the actual id, eg 5, the report is deleted. But when I put $report_id, nothing is deleted.

Please could somebody tell me what correction I have to make to get this to work ?

Here is the code (EDIT: This is the fixed code. I added the hidden field in the form at the bottom, among a few other small changes (like taking out the extra form tag)):


<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN"
<html xmlns="" xml:lang="en" lang="en">
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Football Central - Remove a Report</title>
<h2>Football Central - Remove a News Report</h2>


//Assign variables from admin_reports.php using $_GET
$report_id = $_GET['id'];

if (isset($_POST['submit'])) {
if ($_POST['confirm'] == 'Yes') {

$report_id = $_POST['id'];
// Delete the image file from the server
@unlink(IMAGE_UPLOADPATH . $image);

// Connect to the database
$dbc = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME)
or die("Unable to connect to the database.");

// Delete the score data from the database
$query = "DELETE FROM news_reports WHERE report_id = '".$report_id."' LIMIT 1"
or die("mysql_query failed - Error: " . mysqli_error());

mysqli_query($dbc, $query) or die("mysql_query failed - Error: " . mysqli_error());

//Display form to confirm delete
echo '<p>Are you sure you want to delete the news report?</p>';
echo '<form method="post" action="removereport.php">';
echo '<input type="radio" name="confirm" value="Yes" /> Yes ';
echo '<input type="radio" name="confirm" value="No" checked="checked" /> No <br />';
echo '<input type="hidden" name="id" value="' . $report_id . '" />';
echo '<input type="submit" value="Submit" name="submit" />';
echo '</form>';

echo '<p><a href="admin_reports.php">&lt;&lt; Back to admin reports page</a></p>';


Answer Source

You are sending the form with post method and retrieving it with get. That can be the source of the problem.

Also you are not sending the id parameter so, there won't be any value for $_get[id] nor $_post[id]

Recommended from our users: Dynamic Network Monitoring from WhatsUp Gold from IPSwitch. Free Download