"supplied argument is not a valid MySQL result resource" error

[heaps21]

Hi, i am getting this error with the following code:

$query = "SELECT title, firstname, surname, address1, address2, address3, postcode,"
."tel, mobile, email"
."FROM user"
."WHERE username='$valid_user'";

us.skill_id=s.skill_id"

$result=mysql_query($query);
$num_rows=mysql_num_rows($result);
$row=mysql_fetch_array($result);

The error is referring ot the last 2 lines of this script but I have absolutely no idea y... can anyone help me?

Cheers,
Andy.

[piz]

Try this:

[...]
$result=mysql_query($query);
echo mysql_error();
[...]

[Nightfire]

us.skill_id=s.skill_id"


That line is the fault. What's it meant to do? You have a double quote there and a missing semi-colon.

If it's meant to be in the query, then you'll need to do

$query = "SELECT title, firstname, surname, address1, address2, address3, postcode,"
."tel, mobile, email"
."FROM user"
."WHERE username='$valid_user'" 

."AND us.skill_id=s.skill_id";

$result=mysql_query($query);
$num_rows=mysql_num_rows($result);
$row=mysql_fetch_array($result); 

This should've also been posted in the mysql forum

[heaps21]

sorry, that wasn't meant to be in the query, mustn't have copied it correctly! Will post to the mySQL forum anyway, cheers!