# Check for similar value with SQL

I've two table (t1 and t2) with 3 identical integer columns: c1, c2 and c3. I want to count how many value in t1 are in t2.

SELECT count(t1.value) FROM t1 INNER JOIN t2 ON  (
t1.c1 = t2.c1 OR t1.c1 = t2.c2 OR t1.c1 = t2.c3 OR
t1.c2 = t2.c1 OR t1.c2 = t2.c2 OR t1.c2 = t2.c3 OR
t1.c3 = t2.c1 OR t1.c3 = t2.c2 OR t1.c3 = t2.c3
)


It doesn't seems a good way to write it (I'll have to add some columns). Is there a better solution to write it without enumerated any possibilities?

I'm using MySQL version 5.6.

• If t1 contains c1 = 1, c2 = 1, and c3 = 2 and the t2 has the values 1 and 2 somewhere, should that count as 1, 2, or 3 in the final total? Commented Feb 25, 2014 at 15:46
• I forget : t1.c1, t1.c2 and t1.c3 are all different. Idem for t2. So your example @rolfl should count 2. Commented Feb 25, 2014 at 15:52

Your question is not very clear... but, the way I understand it is:

Collect all the unique values in t1, and count how many of those unique values appear in t2.

Interesting problem.... instead of a straight join with all the or conditions, which may lead to an internal cross-product (thousands of joins and results to run comparisons on), I would state the logic as a couple of subselects ... which represent the two sets of data... the unique values in t1, and the unique values in t2.

Note, the 'union' operator does a distinct as part of the union....

select count(*)
from
(
select c1 as val from t1
union
select c2 as val from t1
union
select c3 as val from t1
) as t1vals,
(
select c1 as val from t2
union
select c2 as val from t2
union
select c3 as val from t2
) as t2vals
where t1vals.val = t2vals.val


The code looks nicer this way, but requires scanning each table three times (which I think will be better than the potentially thousands of times it may have to happen with your query......

## I have put together an sqlfiddle for this

• It works fine, this is a good idea. It´s quite better than enumerate like I done, but it´s not factorized yet (maybe it´s not possible). Thanks. Commented Feb 26, 2014 at 8:45