PHP, MySQL - How to correctly pass a variable in to a link

Question

So what I have is:

$id = $_POST['id'];
$loann = $_POST['loann'];
$dater = $_POST['dater'];
$apaid = $_POST['apaid'];

... and:

$result = mysql_query("UPDATE _loan SET loana='$loann', dater = CONCAT_WS(',', dater,  '$dater' ), apaid = CONCAT_WS(',', apaid , '$apaid' ) WHERE id=$id");

    echo $id;

Everything works fine now the problem shows when I tray to pass the value of $id into:

echo ('<meta http-equiv="refresh" content="1;url=view_details.php?id=$id">');

What I would get in the browser would be:

mysite.net/view_details.php?id=$id

Instead of:

mysite.net/view_details.php?id=133

Any help is greatly appreciated.

Why not use the PHP `header()` function for redirection ? http://php.net/manual/en/function.header.php — Maximus2012, Nov 15 '16 at 17:04
Can you update the question with the code where you used `header()` to see why it did not work? This seems to be a pretty simple problem so it should work. — Maximus2012, Nov 15 '16 at 17:06

score 5 · Accepted Answer · edited May 23 '17 at 12:30

5

You're using a single-quoted string, which doesn't expand variables. Either use a double-quoted string:

"<meta http-equiv=\"refresh\" content=\"1;url=view_details.php?id=$id\">"

or concatenate the strings:

'<meta http-equiv="refresh" content="1;url=view_details.php?id=' . $id . '">'

Side note: Your code is vulnerable to SQL injection. You should start reading here and here.

edited May 23 '17 at 12:30

Community

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answered Nov 15 '16 at 17:06

David

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PHP, MySQL - How to correctly pass a variable in to a link

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