Word Finder regular expression, PHP & mySQL - php

Question

I have a query that I'm trying to run against a database of words. The query must return a word that contains another word, and letters given. It is in PHP and mySQL.
For example:
Word Given: Cruel
Letters Given: abcdty
In the database, I need to find the word "Cruelty" based on the letters given, and the word given. It needs to works both ways. So if I had "atni" for letters, "Anticruel" would appear if it existed in the database.
I have it half working but the result given is not correct:
SELECT word
FROM words
WHERE LOCATE( "cruel", word ) >0
AND word != "cruel"
AND word
REGEXP '[ybilteh]'
The result set from this query:
"anticruelty"
"crueler"
"cruelest"
"crueller"
"cruellest"
"cruelly"
"cruelness"
"cruelnesses"
"cruelties"
"cruelty"
Update!!!
Thanks to Benjamin Morel, this is getting much closer.
This query:
SELECT word
FROM words
WHERE LOCATE( "t", word ) >0
AND word != "t"
AND word
REGEXP '^[ybilteh]*t[ybilteh]*$'
LIMIT 0 , 30
Finds words correctly. But also includes words with double letters. Such as "Beet". When only 1 "e" is available.

Answer 1

Try this one:
SELECT word
FROM words
WHERE word REGEXP '^[ybilteh]*cruel[ybilteh]*$'
AND word != 'cruel';
UPDATE: let's go refining with PHP, what about this?
$word = 'cruel';
$letters = 'ybilteh';
$items = array("anticruelty", "crueler", "cruelest",
"crueller", "cruellest", "cruelly", "cruelness",
"cruelnesses", "cruelties", "cruelty");
$letters = str_split($letters);
foreach ($items as $item) {
$list = $letters;
// remove the original word (once)
$thisItem = preg_replace("/$word/", '', $item, 1);
for ($i=0; $i<strlen($thisItem); $i++) {
$index = array_search($thisItem[$i], $list);
if ($index === false) {
continue 2; // letter not available
}
unset($list[$index]); // remove the letter from the list
}
echo "$item\n"; // passed!
}
Returns: cruelly, cruelty
You might probably find a better/simpler approach, but that should do the trick!