List code not working

sfmnetsys

What's wrong with this code?

<?php

// includes
include("../../includes/config.php");
include("../../includes/functions.php");
include ("../../includes/connect.php");

// generate and execute query
$list_computername_sql = ("SELECT cname,machid FROM inventory WHERE clientid='$clientid'");
$list_computername_result = mysql_query($list_computername_sql);
$numrows = mysql_num_rows($list_computername_result);
if ($numrows>0) {
while ($list = mysql_fetch_array($list_computername_result)) {
echo "<li><a href=\"openlinq.php?module=inventory&clientid=$clientid&machid=".$list["machid"]."\">".$list["cname"]." </a> </li>";
}

}else{
echo"There are no computers available at this time.";
}

To test the code I use a URL with a valid clientid select (ie www.mysite.com?client=1). I get the error:
Undefined variable: clientid on the select statement line. When I add substitute a numeral for $clientid the code works. This is quite frustrating.

Any thoughts?

Thanks.

Anon

The variable you are passing on the querystring is "Client" and not "clientid" mabe changing the names to match will solve it !!

sfmnetsys

Correction. The url would be www.mysite.com/index.php?clientid=1

Anon

Not sure if this will do the trick for you, but do the single quotes around $client_id in your sql string make a difference?

Try:
$list_computername_sql = ("SELECT cname,machid FROM inventory WHERE clientid=$clientid");

zalath

Hmm, is it an error or a warning? if it's a warning, try and put "error_reporting(E_ERROR);" to the top of your code, and see if it helps.

-L-

Anon

It's not a warning, but an error. Single quoations are necessary in order to properly parse. I tried using single, double, and none (quoatation marks)and none of them worked.

wchris

I think it´s simple. You have to write:

while ($list == mysql_fetch_array($list_computername_result)) { ...... }

Anon

At the top, try
$clientid = $_GET['clientid']; // PHP 4.1.0 and above
or
$clientid = $HTTP_GET_VARS['clientid']; // before PHP 4.1.0

Diego

Anon

Nope, '=' is correct here.

(But why is the SQL string is in ()?)

My money is on Lauri Kotilainen's suggestion. Check with error_reporting() to see if that makes the warning go away, and if it does I'd recommend replacing $clientid in the string with $_GET['clientid'] (I'm guessing that since you're getting the warning in the first place you have a version of PHP>=4.1.)

wchris

Oh, I haven´t to write == to compare 2 elements? Why and when I have to make == ?

Anon

You use == when you want to compare two elements.

Steve doesn't want to compare two elements, he wants to set $list to an array from the recordset.

sfmnetsys

Thanks. This worked.