Question

我正在尝试按照他们提交给2个不同表格的条目数量对用户进行排名。

表格gvr：

rid | jid 
---------------
1     54
2     54
3     54
4     75
5     75

表格gos：

sid | jid
---------------
1     54
2     54
3     75
4     75
5     23
6     23

期望的结果：

jid | overall_cnt | gvr_cnt | gos_cnt
----------------------------------
54    5            3          2
75    4            2          2
23    2            0          2

我有：

(SELECT jid, count(*) gvr_count
FROM gvr 
WHERE jid IS NOT NULL
GROUP BY jid)
UNION ALL 
(SELECT jid, count(*) gos_count
FROM gos
WHERE jid IS NOT NULL
GROUP BY jid)

但这是非常不正确的。我一直在寻找类似于我的情况的其他帖子，但找不到任何太多的价值。我正在考虑将数据操作加载到PHP上，但在一个查询中这样做会很方便。

Answer 1

您的查询非常接近。您需要union all，然后执行group by：

select jid, sum(gvr_count) + sum(gos_count) as Overall_Count,
        sum(gvr_count) as gvr_count, sum(gos_count) as gos_count
from ((SELECT jid, count(*) gvr_count, 0 as gos_count
       FROM gvr 
       WHERE jid IS NOT NULL
       GROUP BY jid
      )
      UNION ALL 
      (SELECT jid, 0 as gvr_count, count(*) gos_count
       FROM gos
       WHERE jid IS NOT NULL
       GROUP BY jid
      )
     ) t
group by jid

我认为这是MySQL中最好的方法，以确保您获得所有＆＃34; jid＆＃34; s，甚至是那些只在一个表中的那些。

Answer 2

我更新了戈登回答这是最好的方法。

 select jid ,sum(gvr_count)+ sum(gos_count) as OverallCount ,
  sum(gvr_count) as gvr_count, sum(gos_count) as gos_count
   from ((SELECT jid, count(*) gvr_count, 0 as gos_count
   FROM gvr 
   WHERE jid IS NOT NULL
   GROUP BY jid
  )
  UNION ALL 
  (SELECT jid, 0 as gvr_count, count(*) gos_count
   FROM gos
   WHERE jid IS NOT NULL
   GROUP BY jid
  )
 ) t
group by jid

Answer 3

SELECT temp.jid,gvr_cnt + gos_cnt as totals,temp.*
FROM
(
SELECT gos1.jid
,(SELECT COUNT(*) from gvr where gvr.jid = gvr1.jid) AS gvr_cnt
,(SELECT COUNT(*) from gos where gos.jid = gos1.jid) AS gos_cnt
FROM gos gos1 left join gvr gvr1 on gos1.jid = gvr1.jid
group by gos1.jid
  ) as temp
group by temp.jid

SQL Fiddle Demo

Answer 4

这可能不完全正确，但我希望你能从这里拿走它：

SELECT innerQuery_1.jid AS jid, 
       (innerQuery_1.gvr_count + innerQuery_2.gos_count) AS overall_cnt, 
       innerQuery_1.gvr_count AS gvr_count,
       innerQuery_2.gos_count AS gos_count
FROM   (SELECT jid, count(*) gvr_count
        FROM   gvr 
        WHERE  jid IS NOT NULL
        GROUP BY jid) AS innerQuery_1,
       (SELECT jid, count(*) gos_count
        FROM   gos
        WHERE jid IS NOT NULL
        GROUP BY jid) AS innerQuery_2
GROUP BY innerQuery_1.jid

计算来自2个不同表的总条目数

4 个答案: