计算来自 2 个不同表的总条目数答案

【问题标题】：Counting the number of overall entries from 2 different tables计算来自 2 个不同表的总条目数
【发布时间】：2013-03-02 21:32:51
【问题描述】：

我正在尝试按我的用户提交到 2 个不同表格的条目数对他们进行排名。

表 gvr：

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

表去：

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。我一直在寻找与我的情况类似的其他帖子，但还没有找到任何有价值的东西。我正在考虑将数据操作加载到 PHP 上，但在一个查询中完成它会很方便。

【问题讨论】：

标签： php mysql sql ranking

【解决方案1】：

我更新了 Gordon 的答案，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

【讨论】：

【解决方案2】：

您的查询非常接近。你想要union all，然后做一个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 中确保获得所有“jid”的最佳方法，即使是那些仅在一个表中的。

【讨论】：

我错过了你在查询后的笔记中提到的条件:)

【解决方案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

【讨论】：

有什么东西会破坏mysql上的这个查询吗？因为它在 sqlfiddle 中工作。但我无法让它在我的系统上运行。

【解决方案4】：

这可能不完全正确，但我希望你能从这里开始：

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

【讨论】：

它没有返回 jid 23