Filter a set of bad words out of a PHP array

Question

I have a PHP array of about 20,000 names, I need to filter through it and remove any name that has the word job, freelance, or project

Answer 1

A regular expression is not really necessary here — it'd likely be faster to use a few stripos calls. (Performance matters on this level because the search occurs for each of the 20,000 names.)

With array_filter, which only keeps elements in the array for which the callback returns true:

$data1 = array_filter($data1, function($el) {
        return stripos($el, 'job') === FALSE
            && stripos($el, 'freelance') === FALSE
            && stripos($el, 'project') === FALSE;
});

---

Here's a more extensible / maintainable version, where the list of bad words can be loaded from an array rather than having to be explicitly denoted in the code:

$data1 = array_filter($data1, function($el) {
        $bad_words = array('job', 'freelance', 'project');
        $word_okay = true;

        foreach ( $bad_words as $bad_word ) {
            if ( stripos($el, $bad_word) !== FALSE ) {
                $word_okay = false;
                break;
            }
        }

        return $word_okay;
});

Answer 2

Use of the preg_match() function and some regular expressions should do the trick; this is what I came up with and it worked fine on my end:

<?php
    $data1=array('JoomlaFreelance','PhillyWebJobs','web2project','cleanname');
    $cleanArray=array();
    $badWords='/(job|freelance|project)/i';
    foreach($data1 as $name) {
        if(!preg_match($badWords,$name)) {
            $cleanArray[]=$name;
        }
    }
    echo(implode($cleanArray,','));
?>

Which returned:

cleanname

Answer 3

Personally, I would do something like this:

$badWords = ['job', 'freelance', 'project'];
$names = ['JoomlaFreelance', 'PhillyWebJobs', 'web2project', 'cleanname'];

// Escape characters with special meaning in regular expressions.
$quotedBadWords = array_map(function($word) {
    return preg_quote($word, '/');
}, $badWords);

// Create the regular expression.
$badWordsRegex = implode('|', $quotedBadWords);

// Filter out any names that match the bad words.
$cleanNames = array_filter($names, function($name) use ($badWordsRegex) {
    return preg_match('/' . $badWordsRegex . '/i', $name) === FALSE;
});

Answer 4

I'd be inclined to use the array_filter function and change the regex to not match on word boundaries

$data1 = array('Phillyfreelance' , 'PhillyWebJobs', 'web2project', 'cleanname');

$cleanArray = array_filter($data1, function($w) { 
     return !preg_match('~(freelance|project|job)~i', $w); 
});

Answer 5

This should be what you want:

if (!preg_match('/(freelance|job|project)/i', $name)) {
    $cleanArray[] = $name;
}