Question

我有一个包含这样数据的表h（好吧，不是真的，这只是一个例子）：

subj_id  q1  q2  q3  q4  q5  q6  num
      1   1   0   0   1   0   0    1
      1   0   0   0   1   0   0    2
      2   1   1   1   1   0   1    1
      2   1   0   0   1   0   0    2
      2   1   1   1   0   0   1    3
      3   0   1   0   0   1   1    1

我想总结每个subj_id的q，得到如下输出：

subj_id   num1   num2   num3  
      1      2      1   null
      2      5      2      4
      3      3   null   null

但我得到以下内容：

subj_id   num1   num2   num3
      1      2      1   null
      1      2      1   null
      2      5      2      4
      2      5      2      4
      2      5      2      4
      3      3   null   null

其中求和的行重复次数与subj_id出现在表格中的次数相同。

我的查询（postgres）如下所示：

select h.subj_id, n1.sum as num1, n2.sum as num2, n3.sum as num3 from ((( h
    left join (select subj_id, q1+q2+q3+q4+q5+q6 as sum from h where num=1) as n1 on h.subj_id=n1.subj_id)
    left join (select subj_id, q1+q2+q3+q4+q5+q6 as sum from h where num=2) as n2 on h.subj_id=n2.subj_id)
    left join (select subj_id, q1+q2+q3+q4+q5+q6 as sum from h where num=3) as n3 on h.subj_id=n3.subj_id) order by h.subj_id

左连接显然不是在这里使用的技巧，但是如何跳过重复的行？

提前致谢！

Answer 1

您的查询可以轻松修改为：

with cte as (
    select subj_id, q1 + q2 + q3 + q4 + q5 + q6 as q, num
    from h
)
select
    subj_id,
    sum(case when num = 1 then q end) as num1,
    sum(case when num = 2 then q end) as num2,
    sum(case when num = 3 then q end) as num3
from cte
group by subj_id
order by subj_id

我认为计划会好得多 - 根本没有加入。

=> sql fiddle demo

简要说明您的查询无效的原因以及如何改进它：

您会收到更多您想要的行，因为您的查询基本上是从表h中选择每一行，然后使用h从表num = 1, 2, 3加入它。您的初始表中有6行，这是合乎逻辑的，您的结果中将有6行;
如果您要进行这样的查询，我强烈建议您在内部查询中使用表的别名。我会帮你理解疑问。在某些情况下，它还可以帮助您避免错误的结果 - 请参阅本主题中的答案 - SQL IN query produces strange result。

-

select
    h.subj_id, n1.sum as num1, n2.sum as num2, n3.sum as num3
from h
    left join (
        select h1.subj_id, h1.q1+h1.q2+h1.q3+h1.q4+h1.q5+h1.q6 as sum
        from h as h1
        where h1.num=1
     ) as n1 on h.subj_id=n1.subj_id
    left join (
        select h2.subj_id, h2.q1+h2.q2+h2.q3+h2.q4+h2.q5+h2.q6 as sum
        from h as h2
        where h2.num=2
    ) as n2 on h.subj_id=n2.subj_id
    left join (
        select h3.subj_id, h3.q1+h3.q2+h3.q3+h3.q4+h3.q5+h3.q6 as sum
        from h as h3
        where h3.num=3
    ) as n3 on h.subj_id=n3.subj_id
order by h.subj_id

连接同一行的实例

1 个答案: