match whole word only without regex

问题:

Since i cant use preg_match (UTF8 support is somehow broken, it works locally but breaks at production) i want to find another way to match word against blacklist. Problem is, i want to search a string for exact match only, not first occurrence of the string.

This is how i do it with preg_match

preg_match('/\b(badword)\b/', strtolower($string));

Example string:

$string = "This is a string containing badwords and one badword";

I want to only match the "badword" (at the end) and not "badwords".

strpos('badword', $string) matches the first one

Any ideas?

回答1:

Assuming you could do some pre-processing, you could use replace all your punctuation marks with white spaces and put everything in lowercase and then either:

• Use strpos with something like so strpos(' badword ', $string) in a while loop to keep on iterating through your entire document;
• Split the string at white spaces and compare each word with a list of bad words you have.

So if you where trying the first option, it would something like so (untested pseudo code)

$documet = body of text to process . ' ' 
$document.replace('!@#$%^&*(),./...', ' ')
$document.toLowerCase()
$arr_badWords = [...]
foreach($word in badwords)
{
    $badwordIndex = strpos(' ' . $word . ' ', $document)
    while(!badWordIndex)
    {
        //
        $badwordIndex = strpos($word, $document)
    }
}

EDIT: As per @jonhopkins suggestion, adding a white space at the end should cater for the scenario where there wanted word is at the end of the document and is not proceeded by a punctuation mark.

回答2:

If you want to mimic the \b modifier of regex you can try something like this:

$offset = 0;
$word = 'badword';
$matched = array();
while(($pos = strpos($string, $word, $offset)) !== false) {
    $leftBoundary = false;
    // If is the first char, it has a boundary on the right
    if ($pos === 0) {
       $leftBoundary = true;
    // Else, if it is on the middle of the string, we must check the previous char
    } elseif ($pos > 0 && in_array($string[$pos-1], array(' ', '-',...)) {
        $leftBoundary = true;
    }

    $rightBoundary = false;
    // If is the last char, it has a boundary on the right
    if ($pos === (strlen($string) - 1)) {
       $rightBoundary = true;
    // Else, if it is on the middle of the string, we must check the next char
    } elseif ($pos < (strlen($string) - 1) && in_array($string[$pos+1], array(' ', '-',...)) {
        $rightBoundary = true;
    }

    // If it has both boundaries, we add the index to the matched ones...
    if ($leftBoundary && $rightBoundary) {
        $matched[] = $pos;
    }

    $offset = $pos + strlen($word);
}

回答3:

You can use strrpos() instead of strpos:

strrpos — Find the position of the last occurrence of a substring in a string

$string = "This is a string containing badwords and one badword";
var_dump(strrpos($string, 'badword'));

Output:

45

回答4:

A simple way to use word boundaries with unicode properties:

preg_match('/(?:^|[^pL\pN_])(badword)(?:[^pL\pN_]|$)/u', $string);

In fact it's much more complicated, have a look at here.