How to exit out of PHP code?

Lauj

I have a page that has html and javascript code for all the layout and graphics. In addition to that, I also have the PHP code that load some data from MySQL database. Now the question is, if the query to load data from the database failed to match any, I want to stop executing the whole PHP block of code but still executing the html and javascript code that comes after the PHP block of code. I'm currently use the exit; to stop the executing of the PHP block of code; however, I found that my menu, which I use javascript, does not show up at the bottom. Instead it has been pushed to show half of the menu bar on the top of the page. Here's the page I'm talking about. If click on the button and the page failed to match anything, you'll see the problem I am facing.

Any suggestion is much appreciated.

Charlie

piersk

[man]die()[/man]??

LordShryku

Depending what you're doing, you can just [man]break[/man].

Lauj

Many thanks for the help. The die(); don't work either. As for the break, it only stops loop not entire code.

Well, perhaps anyone can show me how to rearrange my menu code. Frankly, I am not sure why the menu bar at the bottom wasn't showing when I use the exit();.

ljCharlie

Lauj

See? I don't understand why when using the exit; the whole code of the page doesn't get executed even html and javascript. If I put a exit on top of the php page, any html code or javascript will not get executed. Is this what the exit suppose to do or what? I thought the exit is to exit <?PHP ?> code only.

ljCharlie

laserlight

Remember, the page is parsed by PHP before being sent to the client.

So terminating the script, in this case, means also terminating the rest of the page.

Lauj

Thank you for the explanation. I'll see if there is a work around then.

Thanks!

Charlie