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Home/ Questions/Q 6856063
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Editorial Team
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Editorial Team
Asked: May 27, 20262026-05-27T01:50:14+00:00 2026-05-27T01:50:14+00:00

I have a database that has a table that has different LIKES that people

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I have a database that has a table that has different LIKES that people have. Each like is stored as a different record. The only info I have in each record is userID, likeID.

The search will be based on the current userID that is signed in. So I’m not matching EVERYONE with EVERYONE but instead matching EVERYONE vs 1

So userID 45 might have likeID 3, likeID 6 and likeID 22 in the table (for instance)

What I want to do is have the table return, in descending order userIDs that they match with based on the total likes they match with someone else.

For example, (based on user 45 example above) it would look at userID 1 and see how many of likeID 3, likeID 6 and likeID 22 they have. then it’d go to userID 2 and see how many of likeID 3, likeID 6 and likeID 22 they have…..it’d do this for everyone.

Then it’d show the results for any that have more than 0 matches for those 3 likeIDs:

Matches:

You and UserID 12 share 3 of the same likes.

You and UserID 87 share 3 of the same likes.

You and UserID 22 share 2 of the same likes.

You and UserID 73 share 2 of the smae likes.

You and UserID 71 share 1 of the same likes.

etc…

I hope that explains what I’m trying to do. I don’t think the mySQL query will be too hard but right now I’m baffled at how to do it!

THANKS IN ADVANCE!

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  1. Editorial Team
    Editorial Team
    2026-05-27T01:50:15+00:00Added an answer on May 27, 2026 at 1:50 am
    select
          UL2.userID,
          count(*) as LikeMatches
       from
          UserLikes UL1
             JOIN UserLikes UL2
                on UL1.LikeID = UL2.LikeID
               AND UL1.UserID != SingleUserBasisParameter
       where
          UL1.UserID = SingleUserBasisParameter
       group by
          UL2.UserID
       order by
          2 desc
    

    The “2” in the order by is the ordinal column representing the COUNT(*)

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