[RESOLVED] Variable in PHP

[grmapper]

Hello Forum,

I am a newbie. I have a MySQL data base with a column "user_id"

I want to extract the value in user_id and parse it to a statement in PHP. The statement is as follows:

$url = BASE_URL . "p1r2o3t4!" .$user. "!e5c6t7e8d9.php"; // Define the URL:

since the user_id value is 39, I was hoping I would get a result of p1r2o3t4!39!e5c6t7e8d9.php as the filename part of $url

Infact I get p1r2o3t4!!e5c6t7e8d9.php No "39" between the exclamation points.

Any help would be appreciated. It seemed like it worked a few times and then wouldn't work. I may be just to confused. Thanks in advance for any help.

[rincewind456]

If it's not there then it's not being selected when you run your sql statement that assigns the value to $user.

You should check that a value is being returned from the sql select before this point and handle the error if none is returned.

[grmapper]

Yep, that did it. So much for the forest and the trees concept. Thanks again.