Question

I have a database table with words from a dictionary.

Now I want to select words for an anagram. For example if I give the string SEPIAN it should fetch values like apes, pain, pains, pies, pines, sepia, etc.

For this I used the query

SELECT * FROM words WHERE word REGEXP '^[SEPIAN]{1,6}$'

But this query returns words like anna, essen which have repeated characters not in the supplied string. Eg. anna has two n‘s but there is only one n in the search string SEPIAN.

How can I write my regular expression to achieve this? Also if there are repeated characters in my search string at that time the repeated characters should reflect in the result.

Answer 1

Since MySQL does not support back-referencing capturing groups, the typical solution of (\w).*\1 will not work. This means that any solution given will need to enumerate all possible doubles. Furthermore, as far as I can tell back-references are not valid in look-aheads or look-behinds, and look-aheads and look-behinds are not supported in MySQL.

However, you can split this into two expressions, and use the following query:

SELECT * FROM words
WHERE word REGEXP '^[SEPIAN]{1,6}$'
AND NOT word REGEXP 'S.*?S|E.*?E|P.*?P|I.*?I|A.*?A|N.*?N'

Not very pretty, but it works and it should be fairly efficient as well.

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To support a set limit of repeated characters, use the following pattern for your secondary expression:

A(.*?A){X,}

Where A is your character and X is the number of times it’s allowed.

So if you’re adding another N to your string SEPIANN (for a total of 2 Ns), your query would become:

SELECT * FROM words
WHERE word REGEXP '^[SEPIAN]{1,7}$'
AND NOT word REGEXP 'S.*?S|E.*?E|P.*?P|I.*?I|A.*?A|N(.*?N){2}'