php error reporting??

[__greg]

I just started using error reporting on my scripts because im an idiot and never really thought about it before...

anyways, i keep getting this error, and i'm not really sure what to do to change it....

Notice: Use of undefined constant op - assumed 'op' in /path_to/test/test.php on line 5

Notice: Undefined index: op in /path_to/test/test.php on line 5

.. and here's an example of the code i'm using that causes this error:

if ($_POST[op] == "upload") {

 // perform upload actions
}

if (!$_POST[op]) {
	$display_block = "	<form enctype='multipart/form-data' action='$_SERVER[PHP_SELF]' method='post'>
							<input type='hidden' name='op' value='upload'>
							<input type='hidden' name='MAX_FILE_SIZE' value='2000000'>
							<input type='file' name='cover' size='55'><br />
							<br />
							<input type='submit' name='submit' value='submit'>
						</form>";


}

anyone know how i can fix this? in a php book i read it taught me to use that hidden [op] field as a way to do that... is that incorrect?

[benson]

$_POST['op']

That is it i think.

[__greg]

yeah that is it, but i just don't understand what is wrong with using that. and since apparently there is something wrong with that... what i can do differently to change it

[benson]

Make from you $POST[op] this $POST['op'].
Notice the ' ?
I think that's the problem just change it and see what it does.

[hiroh2k]

I think using strings withouth " " is a bad habit of programmers, php allow that but fires a warning:

Use of undefined constant op - assumed 'op'

PHP think op is a constant and it isn't. Then it change it to 'op' and then start to use it.

[LordShryku]

http://www.php.net/manual/en/language.types.array.php

Why is $foo[bar] wrong?

You should always use quotes around a string literal array index. For example, use $foo['bar'] and not $foo[bar]. But why is $foo[bar] wrong? You might have seen the following syntax in old scripts:

<?php
$foo[bar] = 'enemy';
echo $foo[bar];
// etc
?>

This is wrong, but it works. Then, why is it wrong? The reason is that this code has an undefined constant (bar) rather than a string ('bar' - notice the quotes), and PHP may in future define constants which, unfortunately for your code, have the same name. It works because PHP automatically converts a bare string (an unquoted string which does not correspond to any known symbol) into a string which contains the bare string. For instance, if there is no defined constant named bar, then PHP will substitute in the string 'bar' and use that.

Note: This does not mean to always quote the key. You do not want to quote keys which are constants or variables, as this will prevent PHP from interpreting them. [/quote]

[ahundiak]

Something else to consider:
define('op','Whatever');
$POST[op]; is now the same as
$POST['Whatever']; which is very different from
$_POST['op'];

If you don't use the quotes then creating a define may change the way your code works.

[__greg]

okay, thank you all. I think I understand now.. So I put in the quotes, but it still pops out this error:

Notice: Undefined index: op in C:\sokkit\site\sitename_com\test\test.php on line 5

Notice: Undefined index: op in C:\sokkit\site\sitename_com\test\test.php on line 43

LINE 5:
if ($_POST['op'] == "upload") {

LINE 43:
if (!$_POST['op']) {

[ahundiak]

The code is probably being executed even though the submit button has not pressed. Hence, $_POST['op'] does not have a value.

To test if the data was posted use something like:

if (isset($_POST['submit'])) {
  /* Do all your POST processing */
  echo $_POST['op'];
}