-
I've modified the code to say this:
Code:
function checkLoggedIn(){
return !empty($_SESSION["user_id"]);
}
if(checkLoggedIn() ) {
echo '<p>Still logged in, and that is important.</p>';
}
I've checked it twice, and still get a blank screen. I'm sure this code is good, but I cannot make it work for me.
-
If you're still getting a blank screen, it's likely that there is another parse error somewhere.
Do you use a code editor with syntax highlighting? code intelligence? Such things are quite awesome
-
Traq, I've been working at this for several hours now and should probably let it go for a while. I'm grateful to you for all your help and, although the problem is not yet solved, I have been learning new things at each step as a result of your suggestions.
(Yes, my IDE is quite a good one and does give me quite a lot of feedback on errors of all kinds.)
I have another question that's been bothering me for a long time. I'll post that in a separate thread for a fresh start.
-
No problem. I'll take a look at your script if you like. If you prefer not to share it, that's fine too, of course.
I'll look for your other question.