Question

I am trying to pass a selected checkbox value from one page to another to run a mysql statement on my db.
This is what I have:

HTML

<form method='POST' action='move_compaudit.php'>
<input type='hidden' name='checkbox' value='0'/>
<input type='checkbox' name='checkbox' value='1'/>

PHP (this file called move_compaudit.php)

<?php
  include('include/dbConnection.php');
  $checkbox = isset($_POST['checkbox']) ? 'Set' : 'NotSet';
  //SQL statement
  $query = "SELECT * FROM compaudit;";
  $results = mysqli_query($dbc,$query) or die('Error querying database');
  $row = mysqli_fetch_array($results);
  $query1 = "DELETE FROM compaudit WHERE serial_no = $row[7] AND $checkbox = 'Set'";    
  //Execute prepared MySQL statement
  //$results1 = mysqli_query($dbc,$query1) or die('Error querying database');
  print_r($query);
  print_r($query1);
?>

My badly printed query: I get this everytime, regardless of click or not clicked.

SELECT * FROM compaudit;DELETE FROM compaudit WHERE serial_no = 12345 AND Set = 'Set'

Answer 1

If you are going to keep your HTML as it is, you need to check the actual value of $_POST[‘checkbox’] instead of checking if it is set. Your hidden field guarantees that even if it is not checked, $_POST will still get a value for “checkbox”.

<?php
include('include/dbConnection.php');
$checkbox = ($_POST['checkbox'] == '1') ? 'Set' : 'NotSet';
....