Question

I found some partial help but cannot seem to fully accomplish what I need. I need to be able to do the following:

I need an regular expression to replace any 1 to 3 character words between two words that are longer than 3 characters with a match any expression:

For example:

walk to the beach ==> walk(.*)beach

If the 1 to 3 character word is not preceded by a word that’s longer than 3 characters then I want to translate that 1 to 3 letter word to '<word> ?'

For example:

on the beach ==> on ?the ?beach

The simpler the rule the better (of course, if there’s an alternative more complicated version that’s more performant then I’ll take that as well as I eventually anticipate heavy usage eventually).

This will be used in a PHP context most likely with preg_replace. Thus, if you can put it in that context then even better!

By the way so far I have got the following:

$string = preg_replace('/\s+/', '(.*)', $string); 
$string = preg_replace('/\b(\w{1,3})(\.*)\b/', '${1} ?', $string);

but that results in:

walk to the beach ==> 'walk(.*)to ?beach' 

which is not what I want. 'on the beach' seems to translate correctly.

Answer 1

I think you will need two replacements for that. Let’s start with the first requirement:

$str = preg_replace('/(\w{4,})(?: \w{1,3})* (?=\w{4,})/', '$1(.*)', $str);

Of course, you need to replace those \w (which match letters, digits and underscores) with a character class of what you actually want to treat as a word character.

The second one is a bit tougher, because matches cannot overlap and lookbehinds cannot be of variable length. So we have to run this multiple times in a loop:

do
{
    $str = preg_replace('/^\w{0,3}(?: \w{0,3})* (?!\?)/', '$0?', $str, -1, $count);
} while($count);

Here we match everything from the beginning of the string, as long as it’s only up-to-3-letter words separated by spaces, plus one trailing space (only if it is not already followed by a ?). Then we put all of that back in place, and append a ?.

Update:

After all the talk in the comments, here is an updated solution.

After running the first line, we can assume that the only less-than-3-letter words left will be at the beginning or at the end of the string. All others will have been collapsed to (.*). Since you want to append all spaces between those with ?, you do not even need a loop (in fact these are the only spaces left):

$str = preg_replace('/ /', ' ?', $str);

(Do this right after my first line of code.)

This would give the following two results (in combination with the first line):

let us walk on the beach now go => let ?us ?walk(.*)beach ?now ?go
let us walk on the beach there now go => let ?us ?walk(.*)beach(.*)there ?now ?go