Question

A MySQL query needs the results of a subquery in different places, like this:

SELECT COUNT(*),(SELECT hash FROM sets WHERE ID=1) 
     FROM sets 
     WHERE hash=(SELECT hash FROM sets WHERE ID=1) 
           and XD=2;

Is there a way to avoid the double execution of the subquery (SELECT hash FROM sets WHERE ID=1)?
The result of the subquery always returns an valid hash value.
It is important that the result of the main query also includes the HASH.

First I tried a JOIN like this:

SELECT COUNT(*), m.hash FROM sets s INNER JOIN sets AS m
     WHERE s.hash=m.hash AND id=1 AND xd=2;

If XD=2 doesn’t match a row, the result is:

+----------+------+
| count(*) | HASH |
+----------+------+
|        0 | NULL | 
+----------+------+

Instead of something like (what I need):

+----------+------+
| count(*) | HASH |
+----------+------+
|        0 | 8115e| 
+----------+------+

Any ideas? Please let me know! Thank you in advance for any help.

//Edit:
finally that query only has to count all the entries in an table which has the same hash value like the entry with ID=1 and where XD=2. If no rows matches that (this case happend if XD is set to an other number), so return 0 and simply hash value.

Answer 1

SELECT  SUM(xd = 2), hash
FROM    sets
WHERE   id = 1

If id is a PRIMARY KEY (which I assume it is since your are using a single-record query against it), then you can just drop the SUM:

SELECT  xd = 2 AS cnt, hash
FROM    sets
WHERE   id = 1

Update:

Sorry, got your task wrong.

Try this:

SELECT  si.hash, COUNT(so.hash)
FROM    sets si
LEFT JOIN
        sets so
ON      so.hash = si.hash
        AND so.xd = 2
WHERE   si.id = 1