How to determine page URL for if/else

[Hanscraft]

There are blocks on my Drupal website which are being called in based upon taxonomy terms. However, I'd like to have PHP look at the URL of the page that is within that taxonomy. If the URL is 'homework-help', then the block will not show.

I have tried using the following PHP code -- though it removes the block off all pages:

Code: Select all

if ($path = $_SERVER['REQUEST_URI']== 'homework-help' ) {
return FALSE;
} else {
return DL_block_show($taxonomies, $views);

[Emy]

use preg_match or in_array

[Hanscraft]

So what I should be writing would look like:

Code: Select all

$in_array = '/homework-help';
if ($in_array == 'homework-help' ) {
return FALSE;
} else {
return DL_block_show($taxonomies, $views);

[Emy]

You can even do like this:

Code: Select all

$myArray = array('homework-help', 'AnotherOne', 'YouCanAddMore');

$searchWord = 'homework-help';

if(in_array($searchWord, $myArray)
{
echo 'Done';
}
else
{
echo 'Try Later';
}