WSoD

ct2k7

Hi,

I'm building a script, however, upon execution, the user sees a white/blank page instead of what they are meant to be seeing. How do I correct this?

there are no errors and there is only one notice which can be ignored at this time.

Script:http://pastebin.com/d5e6bd567

TiA

NogDog

No way I'm going to read a couple thousand lines of source code, without any clue as to where the problem is. However, just looking at the top of the file, you do not appear to do anything if $_SESSION['user'] is not set, which means you're going to have an undefined variable a few lines later where you reference $user if it's not set.

But really, this is a good example of why modularization of code either via functions or objects could make things a lot easier to debug, rather than just one "stream of consciousness" script. At this point, without a major rewrite, all I can suggest is sticking in [man]var_dump/man's and/or [man]debug_print_backtrace/man's to find out what values various things have, what your query strings really look like, etc.

ct2k7

I'm awful at breaking this down 😐 I'm not sure where to start. As the script is not standalone, and will be called from the "main" script, which contains session_start, does this affect anything?

j_70

When you say that there are no errors, where are there no errors? What does your apache log say when you run the page?

HalfaBee

If you can write scripts, you should be able to debug them.

put echo() and print_r() as NogDog suggested, you can then find which part of the script is not being executed.

NogDog

Also, if you get a blank browser screen:

Do a "View Source" in the browser and see if there's anything interesting showing up there.
If you can, either change the global PHP configuration (e.g. in php.ini) or at the local directory level via a .htaccess file in order to turn on error_reporting and display_errors, as parse errors will not be affected by those settings in the script itself.

.htaccess:

php_flag display_errors on
php_value error_reporting E_ALL

ct2k7

Parse error: unexpected $End on line 1122

NogDog

That error normally indicates something has not been closed: often a missing "}" to close a code block, but possibly a missing closing quote or other delimiter.

ct2k7

Error is after line 16

http://pastebin.com/m34b3e0dc

My guess is that I'm not closing the if where I should close it?

laserlight

You need to indent your code properly. It will probably help you discover and fix your error.

ct2k7

laserlight;10888558 wrote:
You need to indent your code properly. It will probably help you discover and fix your error.

doing so right now -

I've used !$var

is that correct in PHP?

laserlight

I've used !$var

is that correct in PHP?

It is certainly syntactically correct on its own, but whether it is logically correct or syntactically correct with respect to the rest of your code is another matter.

ct2k7

How would I know?

laserlight

How would I know?

By inspection and testing.

ct2k7

Could you please simplify? (I'm not working in full throttle today)

laserlight

Could you please simplify?

I can't, because I need context to tell you whether !$var makes sense or not.

ct2k7

Updated script is here: http://pastebin.com/m4b41b2c8

Added indentation.

For !$var:

if ($oc == "0"){

$sub = strip_tags($_POST['start_oc']);
$split = strip_tags($_POST['split']);

	if ($sub){
		echo("two.split");
		if ([B]!$split[/B]){
		$error = "Please select the type of cash split you wish to use for your OC.";
		die($error);
	}elseif ($split){