undefined variable? huh?

Anon

Ive been using the following code on several pages and in the past it has worked flawlessly and never generated any warnings..

if(!($upload))
{

$query = "SELECT * FROM RevBookSetup WHERE ID = 'one' ";
...

anyways, if the script is run and no value is passed for $upload, it does this, and if $upload is set , it does this instead. Its always worked before without any warnings showing up. Now, I keep getting this error:

Warning: Undefined variable: upload in /home/idahopag/public_html/witchesrealm/wr/revbook/revbookadmin/revbookadmin.php3 on line 18

Sometimes it shows up, sometimes it doesn't.. and its got me really confused..

any help is greatly appreciated!

Anon

Did you just update php? I did and I got the same errors. Simple solution, just use the isset() function to check if the variable exists.

if(!isset($upload))
{

$query = "SELECT * FROM RevBookSetup WHERE ID = 'one' ";
...

The new versions of php seem to be more strict in the old version variables that weren't set defaulted to either 0 or an empty string, now they give errors. Oh well probably for the best, keeps our code clean.

Anon

Mucho grassias.. or however you spell it.. I didn't update php, but then again, its not my server.. they might have updated tho.. either way, that works like a charm.. I should have thought about using that, but I guess my brain is just too tired from working on this all day...

thanx man! 🙂 🙂

Anon

Heh, no prob, my brain did a triple-backflip too when I ran into this error.

Anon

I am having the same problem. I have put in the isset and still get the undefined variable.
<?

include("../info.php");
if(!isset($trucat))
{
if ($trucat == "")
{$trucat = 'aar';}

$result = mysql_query ("SELECT * FROM products
WHERE name LIKE '$trucat%'");
if ($row = mysql_fetch_array($result)) {
do {
print $row["name"];
print ("<p>");
} while($row = mysql_fetch_array($result));
} else {print "Please select a category from above!";}
}
else {print "Please select a category from above!";}
?>