Passing php-variable with JavaScript and HTML-Form

Question

I am trying to create a Delete-button with a user confirmation-pop-up.

HTML/PHP:

<form method='POST' onclick='return confirmDelete()'>
 <button type='submit'>
 Delete script
 </button>
</form>

JavaScript:

<script type="text/javascript">
function confirmDelete() {
if (confirm("Are you sure?")) {
 window.location.href = 'delete_script.php?<?php echo "$id" ?>';
 return false;
}
else{
 return true;
 }
}
</script>

What it does is that it re-directs me to the correct page, "delete_script.php" but the $id does not appear in the url. The "delete_script.php"-page works with other pages, but I don't know what's wrong with this code. I really think it's possible. Any help would be appreciated!

• synapze: This looks od <?php echo "$id" ?>. If $id is a variable remove the double quotes. You also might want to add a query string name for example 'delete_script.php?id=<?php echo $id ?>'.
• ilarsona: You could just keep it in a $_SESSION[''] varible... that way it's global... everywhere .
• Ja͢ck: Where does $id come from in the first place?
• simeg: I did try to keep it in a $_SESSION variable, but it wouldn't get into the url anyway. And I'd prefer to do it this way. I have excluded alot of code, the $id is the id of a post in my database. This page gets it from $_GET from another page.

Answer 1

You're expected to pass a url-encoded value for the query-string:

<?php echo 'id=' . urlencode($id); ?>

Alternatively, generate the whole url in php:

location.href = <?php echo json_encode('delete_script.php?id=' . urlencode($id)); ?>;

• simeg: The alternative answer worked perfectly! Thank you very much, sir.

Answer 2

As per my comment, change this...

if (confirm("Are you sure?")) {
 window.location.href = 'delete_script.php?<?php echo "$id" ?>';
 return false;
}

to this..

if (confirm("Are you sure?")) {
 window.location.href = 'delete_script.php?id=<?php echo $id ?>';
 return false;
}

• simeg: No difference, thanks though. Just re-directing me to the page without any $id in the url.