Test if a string contains at least 2 words in SQL - sql

Question

I have an sql request like :
SELECT *
FROM table
WHERE lower(title) LIKE lower('%It's a beautiful string i think%')
I need to check if at least 2 words in my string It's a beautiful string i think are contained in my field title... How can i do that ?
For example, if in my field title i have the string I think it's beautiful, This query should return me this object...
Thanks !

Answer 1

You could split your string into a temporary table (say, using something like this: http://ole.michelsen.dk/blog/split-string-to-table-using-transact-sql/) and then do a join, with a count.

Answer 2

You could generate the following SQL statement dynamically:
SELECT title, count(*)
FROM
(
SELECT title
FROM table1
WHERE (' ' + lower(title) + ' ') LIKE lower('% It %')
UNION ALL
SELECT title
FROM table1
WHERE (' ' + lower(title) + ' ') LIKE lower('% s %')
UNION ALL
SELECT title
FROM table1
WHERE (' ' + lower(title) + ' ') LIKE lower('% a %')
UNION ALL
SELECT title
FROM table1
WHERE (' ' + lower(title) + ' ') LIKE lower('% beautiful %')
UNION ALL
SELECT title
FROM table1
WHERE (' ' + lower(title) + ' ') LIKE lower('% string %')
UNION ALL
SELECT title
FROM table1
WHERE (' ' + lower(title) + ' ') LIKE lower('% I %')
UNION ALL
SELECT title
FROM table1
WHERE (' ' + lower(title) + ' ') LIKE lower('% think %')
) AS table2
GROUP BY title
HAVING COUNT(*) >= 2
A stored procedure might be more efficient and you could have the whole job done on the server side.

Answer 3

You could use a function like this
CREATE FUNCTION [dbo].[CheckSentece] (#mainSentence varchar(128), #checkSentence varchar(128))
RETURNS NUMERIC AS
BEGIN
SET #mainSentence=LOWER(#mainSentence)
SET #checkSentence=LOWER(#checkSentence)
DECLARE #pos INT
DECLARE #word varchar(32)
DECLARE #count NUMERIC
SET #count=0
WHILE CHARINDEX(' ', #checkSentence) > 0
BEGIN
SELECT #pos = CHARINDEX(' ', #checkSentence)
SELECT #word = SUBSTRING(#checkSentence, 1, #pos-1)
DECLARE #LEN NUMERIC
//Simple containment check, better to use another charindex loop to check each word from #mainSentence
SET #LEN=(SELECT LEN(REPLACE(#mainSentence,#word,'')))
if (#LEN<LEN(#mainSentence)) SET #count=#count+1
SELECT #checkSentence = SUBSTRING(#checkSentence, #pos+1, LEN(#checkSentence)-#pos)
END
return #count
END
and get the number of words from second sentence contained in the first one