Ranking with LIKE keyword in SQL Server - sql

Question

I have table like
╔═══════════════════╗
║ Title ║
╠═══════════════════╣
║ Blogroll ║
║ Bottom Menu ║
║ Business ║
║ Entertainment ║
║ extend ║
╚═══════════════════╝
and my search criteria is like
WHERE title LIKE '%blogroller%'
obviously I will have no result here but can I find the count where the like clause ended, like here in this case its beyond 8 that made like clause fail??
Any hint would be appreciated.
Thank you

Answer 1

You can do this but only with lots of manual effort:
select title,
(case when title like '%blogroller%' then 10
when title like '%blogrolle%' then 9
when title like '%blogroll%' then 8
. . .
else 0
end) as MatchLen
from table t
order by MatchLen desc;
(Note: in some versions of SQL Server, you might need a subquery to refer to MatchLen.)

Answer 2

You don't have ability to do it at once, so you have to try all options one by one, but you can significantly optimize this process - that's what SQL for.
First, convert pattern into all possible patterns, then just find MAX from all possible LIKEs. Not sure if SQL will be able to really optimize it now, but maybe in future it will.
-- test data
DECLARE #token NVARCHAR(100)
SET #token = 'Blogroller'
DECLARE #titles TABLE (Title NVARCHAR(100))
INSERT #titles VALUES
('Blogroll'),
('Bottom Menu'),
('Business'),
('Entertainment'),
('extend')
-- solution
DECLARE #patterns TABLE (part NVARCHAR(100) PRIMARY KEY, tokenLen int)
WHILE (LEN(#token) > 0)
BEGIN
INSERT #patterns VALUES (#token, LEN(#token))
SET #token = SUBSTRING(#token, 1, LEN(#token) - 1)
END
SELECT MAX(patterns.tokenLen)
FROM #titles titles
INNER JOIN #patterns patterns ON titles.Title LIKE '%' + patterns.part + '%'

Answer 3

If you don't mind declaring a couple of variables and running a simple loop, you can do it this way:
declare #phrase as varchar(100) = 'blogroller'
declare #match_length as int = 0
while len(#phrase) > 0
begin
if (select count(title) from titles where title like '%' + #phrase + '%') > 0
begin
set #match_length = len(#phrase)
break
end
set #phrase = LEFT(#phrase, len(#phrase) - 1)
end
print #match_length

Answer 4

You can use some table with numbers (master..spt_values used in example) to break your search term into all possible shorter terms, and then join on it to find how many characters have matched:
setup:
CREATE TABLE Something (Title NVARCHAR(MAX));
INSERT Something (Title)
VALUES ('Blogroll')
,('Bottom Menu')
,('Business')
,('Entertainment')
,('extend')
query:
DECLARE #SearchTerm NVARCHAR(MAX);
SET #SearchTerm = 'Blogroller';
WITH CTE_SearchBroken AS
(
SELECT LEFT(#SearchTerm, number) BrokenTerm
FROM master..spt_values n
WHERE n.type = 'P' AND n.number <= LEN(#SearchTerm)
)
, CTE_PreliminaryResults AS
(
SELECT *, LEN(BrokenTerm) AS BrokenAt
FROM Something s
LEFT JOIN CTE_SearchBroken b ON s.Title LIKE '%' + b.BrokenTerm + '%'
)
SELECT Title
, MAX(BrokenAt) AS BrokenAt
, CASE WHEN LEN(#SearchTerm) = MAX(BrokenAt) THEN 1 ELSE 0 END AS Found
FROM CTE_PreliminaryResults
GROUP BY Title

Answer 5

Based on Gordon's answer, you can write a function that automatically loops through all the possibilities. Also similar to Lanorkin's answer; his might have better performance because it makes good use of inner join; you may want to do some tests there.
CREATE FUNCTION dbo.MatchLen(#needle AS varchar(max), #haystack AS varchar(max)) RETURNS int
AS
BEGIN
DECLARE #len AS int = LEN(#needle)
WHILE #len > 0 AND #haystack NOT LIKE '%' + REPLACE(REPLACE(REPLACE(LEFT(#needle, #len), '[', '[[]'), '%', '[%]'), '_', '[_]') + '%'
SET #len = #len - 1
RETURN #len
END
Example:
SELECT dbo.MatchLen('blogroller', 'This blog has no blogroll, sir!')
returns 8.

Answer 6

Please try:
select
*,
LEN(Title) MatchLength
From YourTable
where 'Blogroller' like Title+'%'
OR
select
*,
LEN(Title) MatchLength
From YourTable
where 'Blogroller' like '%'+Title+'%'