mysql: matching a query letter by letter - php

Question

I am in the process of learning MySQL and querying, and right now working with PHP to begin with.
For learning purposes I chose a small anagram solver kind of project to begin with.
I found a very old English language word list on the internet freely available to use as the DB.
I tried querying, find in set and full-text search matching but failed.
How can I:
Match a result letter by letter?
For example, let's say that I have the letters S-L-A-O-G to match against the database entry.
Since I have a large database which surely contains many words, I want to have in return of the query:
lag
goal
goals
slag
log
... and so on.
Without having any other results which might have a letter used twice.
How would I solve this with SQL?
Thank you very much for your time.

Answer 1

$str_search = 'SLAOG';
SELECT word
FROM table_name
WHERE word REGEXP '^[{$str_search}]+$' # '^[SLAOG]+$'
// Filter the results in php afterwards
// Loop START
$arr = array();
for($i = 0; $i < strlen($row->word); $i++) {
$h = substr($str_search, $i, 0);
preg_match_all("/{$h}/", $row->word, $arr_matches);
preg_match_all("/{$h}/", $str_search, $arr_matches2);
if (count($arr_matches[0]) > count($arr_matches2[0]))
FALSE; // Amount doesn't add up
}
// Loop END
Basicly run a REGEXP on given words and filter result based on how many occurencies the word compared with the search word.
The REGEXP checks all columns, from beginning to end, with a combination of given words. This may result in more rows then you need, but it will give a nice filter nonetheless.
The loop part is to filter words where a letter is used more times then in the search string. I run a preg_match_all() on each letter in found the word and the search word to check the amount of occurencies, and compare them with count().

Answer 2

If you want a quick and dirty solution....
Split the word you're trying to get anagrams for into individual letters. Assign each letter an individual prime number value, and multiply them all together; eg:
C - 2
A - 3
T - 5
For a total of 30
Then step through your dictionary list, and do the same operation on each word in that. If your target word's value is divisible exactly by the dictionary word's value, then you know that the dictionary word has only letters that occur in your target word.
You can speed it up by pre-calculating the dictionary values, and then querying for just the right values:
SELECT * FROM dictionary WHERE ($searchWordTotal % wordTotal) = 0
(searchWordTotal is the total for the word you're looking for, and wordTotal is the one from the database)
I should get around to writing this properly one of these days....

Answer 3

since you only want words with the letters given, and no others, but you dont need to use all the letters, then i suggest logic like this:
* take your candidate word,
* do a string replace of the first occurrence of each letter in your match set,
* set the new value to null
* then finally wrap all that in a strlength to see if there are any characters left.
you can do all that in sql - but a little procedure will probably look more familiar to most coders.