Question

I’m having trouble coming up with a working query for this situation…

Table:

[ matchID ]  [ User ] [ team ]
     1          10       1
     1          77       2
     2          10       1
     2          77       1
     3          10       2
     4          10       1

Notes:

• sometimes the opponent is unknown, for example in matches 2, 3, and 4
• sometimes a team has only one person on a side, for example match 1
• sometimes a team has n persons on a side, for example match 2

Desired Query Results:

A list of matchIDs for a given user where the opponent is unknown.

My first attempt was:

SELECT matchID FROM table
GROUP BY matchID
HAVING COUNT(matchID) = 1

but then I realized that I wasn’t counting matches where there are teammates but unknown opponents, such as match 2 above.

In plain English, I need to get all matchIDs where:

• involving a given user <- edited to add
• everyone is on the same team
• OR there is only one person listed

but I’m not sure how to do that efficiently in one query. Help?

Answer 1

SELECT MatchID
  FROM Table
 GROUP BY MatchID
HAVING COUNT(DISTINCT Team) = 1;

You seem to need those matches where only one team is recorded in the table, and this query gives you those matches.

If the match must involve a specific user (10 for the sample data), then:

SELECT t1.MatchID
  FROM Table AS t1
 WHERE EXISTS(SELECT *
                FROM Table AS t2
               WHERE t2.UserID = 10 AND t1.MatchID = t2.MatchID)
 GROUP BY t1.MatchID
HAVING COUNT(DISTINCT t1.Team) = 1;

You can probably do that as an (inner) self-join:

SELECT t1.MatchID
  FROM Table AS t1
  JOIN Table AS t2 ON t1.MatchID = t2.MatchID AND t2.UserID = 10
 GROUP BY t1.MatchID
HAVING COUNT(DISTINCT t1.Team) = 1;

Or, perhaps a little more clearly:

SELECT t1.MatchID
  FROM Table AS t1
  JOIN (SELECT MatchID FROM Table AS t2 WHERE t2.UserID = 10) AS t2
    ON t1.MatchID = t2.MatchID
 GROUP BY t1.MatchID
HAVING COUNT(DISTINCT t1.Team) = 1;

The sub-select picks those matches where UserID 10 played; the rest works as before.