[RESOLVED] Advice on extracting words from a string

[timski72]

Hello,

I would like to extract individual words from a string. I want to ignore any punctuation marks.

At the moment, I am using the following:

 $words = preg_replace("/[^a-zA-Z\s]/", "", $words);
 $resultingWords = explode(" ", $words);

This works to a certain extent but if there are two spaces inbetween a word then the space is 'exploded' as a word if you get what I mean.

For example "This, for example, (note that was 2 spaces just now) would result in the following array!" - see attachment.

So punctuation has been stripped out as required and only characters left, but where there were 2 blanks one of them has been interpreted as part of the array.

Is there a way I can avoid this? If i strip out all blanks I end up with just one long string. Probably there is a better way of approaching this?

I'd appreciate any suggestions.

Thanks,
Tim.

[bradgrafelman]

I would say you're going about it the wrong way - instead of trying to replace things you don't want, why not look for things you do want? Quick example:

$str = "This, for example,  (note that was 2 spaces just now) would result in the following array!";
preg_match_all('/\b[a-z]+\b/i', $str, $matches);
$resultingWords = $matches[0];

$resultingWords would then be:

Array
(
    [0] => Array
        (
            [0] => This
            [1] => for
            [2] => example
            [3] => note
            [4] => that
            [5] => was
            [6] => spaces
            [7] => just
            [8] => now
            [9] => would
            [10] => result
            [11] => in
            [12] => the
            [13] => following
            [14] => array
        )

)

[timski72]

Thanks bradgrafelman!

[dalecosp]

Yay!

To the OP: this might be a good place to hit that "RESOLVED" button. 🙂