[RESOLVED] HELP!!! query problem

hello,

i am trying to display certain photos from a database, depending on what user is logged in.

now the login part is working , but when i use a query to select the photos belonging to that user, nothing gets displayed.

this is the query:

$sql1 = "select plus_signup.person_ID from plus_signup where plus_signup.userid = ".$session[userid];

now when i echo this on the screen, nothing is returned..
it should return a 1 as the person_ID is 1 for the name sander

when i echo $session[userid] the name sander gets displayed so i know the $session has the right value

anybody any ideas?

$sql1 = "select plus_signup.person_ID 
    from plus_signup 
    where plus_signup.userid = '".$session['userid'] . "'";

I added a few ' signs. If userid is a int then you don't have to use the ' signs before and after the $session[...] part.

Exactly what is echoed if you echo $sql1? Does the query work if you run it from phpmyadmin or any simular program?

i tried your query.

now when i echo $sqll i get

sqll = select plus_signup.person_ID from plus_signup where plus_signup.userid = 'sander'

looks good, but still no result in the webpage.

the funny thing is, that is does work in PHPmyadmin when i enter the query directly like this

SELECT plus_signup.person_ID FROM plus_signup WHERE plus_signup.userid = 'sander'

the result is then exactly what it should be

As I see it there are two possibilities:

You don't have any match in the database. Check it by using PHPMyAdmin.
There is something wrong with the rest of your code. Please post that code as well.

ok there was a mistake in the code.. i fixed it, but now i get a response i dont recognize.

$sqll = "select plus_signup.person_ID from plus_signup where plus_signup.userid = '".$session['userid'] . "'";

echo " sqll = $sqll<br>";

$user = mysql_query($sqll);

echo "username = $user<br>";

$sql = "SELECT * FROM ".$table." where person_ID = ".$user." ORDER BY image_id";

$result = mysql_query ($sql, $conn);

now what gets echod on the screen is :

sander
sqll = select plus_signup.person_ID from plus_signup where plus_signup.userid = 'sander'
username = Resource id #5

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /home/smashed/public_html/signup/welcome.php on line 163

i have no idea what Resource id#5 is...