PHP: Check URL against array of badwords, and output any found - php

Question

I have a database which contains the results of logged network traffic, and I'm building a simple log viewer frontend in PHP. I need to check each URL and see if it contains any of the 'bad words' in an external 'badwords.txt' file, then echo which ones, and color the table row red (the easy bit!)
Thus far I have loaded the badwords.txt file into an array, and fetched the URL from the database. Here is the portion of my code where I am trying to get a positively identified 'bad' url. I ideally want to output which badwords were found, but have simplified everything to just try and get it to work for now.
// Load the badwords file into an array
$words = file('badwords.txt');
//$row[3] is the URL fetched from the database
$testURL = $row[3];
foreach ($words as $phrase) {
if (strrpos($testURL, $phrase)) {
echo "FOUND";
}
}
This is not working for me, and never outputs FOUND, even when the url definitely contains a bad word. I have checked that my $words array is populated correctly with all the badwords, and I have checked that the $testURL is not empty etc.
Can anyone help please? :) I'd be really grateful for any assistance - I have read through so many StackExchange posts on similar topics, but none seem to work for my case.
Thank you!

Answer 1

Words array contains newline symbol, so strpos never works. Remove them
$words = file('badwords.txt', FILE_IGNORE_NEW_LINES);

Answer 2

Since strrpos() returns numeric position of needle, and boolean false, so code should be
<?php
// Load the badwords file into an array
$words = file('words.txt', FILE_IGNORE_NEW_LINES);
//$row[3] is the URL fetched from the database
$testURL = row[3];
foreach ($words as $phrase) {
if (strrpos($testURL, $phrase)>-1) {
echo "FOUND";
}
}