0

我有以下查询，它正在查找计数：

stage_2 IS NULL stage_1 等于 1 member_group 是 5 按 rel_id 分组

至少有 2 行符合上述要求。

这工作正常。它输出正确的结果 6。（在我的小提琴中）。

SELECT COUNT(*) AS count_result
FROM (
    SELECT sub.entry_id
    FROM exp_judging AS jud
    INNER JOIN exp_submissions AS sub ON jud.rel_id = sub.id 
    WHERE jud.stage_2 IS NULL
    AND jud.stage_1 = 1 
    AND sub.member_group = 5
    GROUP BY jud.rel_id
    HAVING COUNT(*) >= 2
) AS a

问题是，如果其中一个评判行在“stage_2”中输入了分数，它会继续计算这个 rel_id，因为在 stage_2 中仍有超过 2 行具有 NULL。

我想要做的是检查上述所有内容是否正确，但同时，judge_id 为“14”的行在 stage_2 中是否为 NULL。

我怎样才能做到这一点？

这是我的表和当前查询的 SQL Fiddle - http://sqlfiddle.com/#!2/e5ee5/1

4

1 回答 1

0

考虑如下中间结果...

  SELECT s.id
      , s.member_group
      , j.judge_id
      , j.stage_1
      , j.stage_2
   FROM exp_judging j
   JOIN exp_submissions s
     ON s.id = j.rel_id
    AND j.stage_1 = 1
    AND s.member_group = 5
  ORDER
     BY j.rel_id
      , j.judge_id;

 +----+--------------+----------+---------+---------+
 | id | member_group | judge_id | stage_1 | stage_2 |
 +----+--------------+----------+---------+---------+
 | 70 |            5 |       14 |       1 |       5 |<-- exclude because [14,5]
 | 70 |            5 |       16 |       1 |    NULL |<-- exclude because of [14,5] above
 | 73 |            5 |       14 |       1 |       5 |<-- exclude because [14,5]
 | 73 |            5 |       15 |       1 |    NULL |<-- exclude because [14,5] above
 | 73 |            5 |       16 |       1 |    NULL |<-- exclude because [14,5] above
 | 75 |            5 |       15 |       1 |    NULL |
 | 75 |            5 |       16 |       1 |    NULL |
 | 77 |            5 |       15 |       1 |    NULL |
 | 77 |            5 |       16 |       1 |    NULL |
 | 78 |            5 |       16 |       1 |    NULL |
 | 79 |            5 |       14 |       1 |    NULL |
 | 80 |            5 |       14 |       1 |    NULL |
 | 80 |            5 |       15 |       1 |    NULL |
 | 80 |            5 |       16 |       1 |    NULL |
 | 81 |            5 |       16 |       1 |    NULL |
 | 83 |            5 |       14 |       1 |    NULL |
 | 83 |            5 |       15 |       1 |    NULL |
 +----+--------------+----------+---------+---------+

要排除突出显示的行，我们可以这样做......

  SELECT s.id
      , s.member_group
      , j.judge_id
      , j.stage_1
      , j.stage_2
   FROM exp_judging j
   JOIN exp_judging x
     ON x.rel_id = j.rel_id
    AND x.judge_id = 14
    AND x.stage_2 IS NULL
   JOIN exp_submissions s
     ON s.id = j.rel_id
    AND j.stage_1 = 1
    AND s.member_group = 5
  GROUP
     BY j.rel_id
      , j.judge_id;

 +----+--------------+----------+---------+---------+
 | id | member_group | judge_id | stage_1 | stage_2 |
 +----+--------------+----------+---------+---------+
 | 75 |            5 |       15 |       1 |    NULL |
 | 75 |            5 |       16 |       1 |    NULL |
 | 77 |            5 |       15 |       1 |    NULL |
 | 77 |            5 |       16 |       1 |    NULL |
 | 78 |            5 |       16 |       1 |    NULL |
 | 79 |            5 |       14 |       1 |    NULL |
 | 80 |            5 |       14 |       1 |    NULL |
 | 80 |            5 |       15 |       1 |    NULL |
 | 80 |            5 |       16 |       1 |    NULL |
 | 81 |            5 |       16 |       1 |    NULL |
 | 83 |            5 |       14 |       1 |    NULL |
 | 83 |            5 |       15 |       1 |    NULL |
 +----+--------------+----------+---------+---------+

从这里开始，剩下的步骤是微不足道的......

 SELECT s.id
      , COUNT(*)
   FROM exp_judging j
   JOIN exp_judging x
     ON x.rel_id = j.rel_id
    AND x.judge_id = 14
    AND x.stage_2 IS NULL
   JOIN exp_submissions s
     ON s.id = j.rel_id
    AND j.stage_1 = 1
    AND s.member_group = 5
  GROUP
     BY j.rel_id;

+----+----------+
| id | COUNT(*) |
+----+----------+
| 75 |        2 |
| 77 |        2 |
| 78 |        1 |
| 79 |        1 |
| 80 |        3 |
| 81 |        1 |
| 83 |        2 |
+----+----------+
于 2013-10-17T09:17:37.440 回答