Wil — Tue, 04 Dec 2001 18:27:00 +0000

Hi Mark

I wasn't claiming that all variables weren't variants, what I was saying was that he hadn't of defined what kind of data strcuture 's' is. Is it a scalar, array, hash, syscall etc.

Mark Hensler — Tue, 04 Dec 2001 17:11:04 +0000

By default, all variables are variant. You can, however, type cast them. But you'll rarely ever see anyone do that in PHP. Examples:

<?php
// make a dummy var
$tmp= \"1234\";

// let's type cast
$myString = (string) $tmp;
$myInt = (int) $tmp;
$myBool = (bool) $tmp;

// let's settype()
$myString = settype($tmp, \"string\");
$myInt = settype($tmp, \"int\");
$myBool = settype($tmp, \"bool\");
?>

PHP Docs: type-casting, type-juggling, settype()

Wil — Tue, 04 Dec 2001 15:21:24 +0000

A lot more of your script would be very useful in helping you with your problems.

You know this variable you are trying to define? Does the variable get defined from a database? A much better way to write that would be:

if (!$s) 
{
 $s = 1;
}

'
Please note that I come from a Perl background and not a PHP one, although both language do share common properties.

Surely you have to state what kind of data structure 's' is? Is it a scalar, array, hash, monkey, elephant?

For your second question, hmm. I think I'll need more information again. How are you preparing and executing this? A few notes, however, the line:

"$sql = "UPDATE TABLE products SET"

should be replaced with:

$sql = "UPDATE products SET

'
You don't need TABLE there. In fact, that could actually be your error.

Put your last statement on a seperate line. It looks cleaner, and you can also enclose that with single quotation marks too.

...
WHERE
Product_ID = '$id'

Cheers