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Home/ Questions/Q 3216474
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Editorial Team
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Editorial Team
Asked: May 17, 20262026-05-17T15:17:55+00:00 2026-05-17T15:17:55+00:00

I am currently using the following query to get some numbers: SELECT gid, count(gid),

  • 0

I am currently using the following query to get some numbers:

SELECT gid, count(gid), (SELECT cou FROM size WHERE gid = infor.gid)       
FROM infor 
WHERE id==4325 
GROUP BY gid;

The output I am getting at my current stage is the following:

+----------+-----------------+---------------------------------------------------------------+
| gid      | count(gid)      | (SELECT gid FROM size WHERE gid=infor.gid)                    |
+----------+-----------------+---------------------------------------------------------------+
|       19 |               1 |                                                            19 | 
|       27 |               4 |                                                            27 | 
|      556 |               1 |                                                           556 | 
+----------+-----------------+---------------------------------------------------------------+

I am trying to calculate the weighted average i.e.

(1*19+4*27+1*556)/(19+27+556)

Is there a way to do this using a single query?

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  1. Editorial Team
    Editorial Team
    2026-05-17T15:17:56+00:00Added an answer on May 17, 2026 at 3:17 pm

    Use:

    SELECT SUM(x.num * x.gid) / SUM(x.cou)
      FROM (SELECT i.gid,
                   COUNT(i.gid) AS num,
                   s.cou
              FROM infor i
         LEFT JOIN SIZE s ON s.gid = i.gid
             WHERE i.id = 4325
          GROUP BY i.gid) x
    
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