这是我的情况。
PayTable
+-------+------+--------------+
| Craft | Job | sequence |
+-------+------+--------------+
| 400 | 1 | 1 |
+-------+------+--------------+
| 401 | 2 | 2 |
+-------+------+--------------+
| 5501 | 2 | 3 |
+-------+------+--------------+
Jobs
+-------+------+------+
| Job | CraftTemplate |
+-------+------+------+
| 1 | 1 |
+-------+------+------+
| 2 | 1 |
+-------+------+------+
Pay Template
+-------+--+
| Template |
+-------+--+
| 1 |
+-------+--+
PayCraftTemplate
+-------+------+---------+
| PayTemplate | Craft |
+-------+------+---------+
| 1 | 400 |
+-------+------+---------+
| 1 | 401 |
+-------+------+---------+我需要做的是从PayTable中找到PayCraftTemplate中不存在的所有工艺品。作为一种反联接模式,这看起来非常简单,但是我似乎不能正确地返回数据。
加入链接为:
PayTable INNER JOIN Jobs by Job -> Job
Jobs LEFT OUTER JOIN Pay Template by CraftTemplate -> Template
Pay Template LEFT OUTER JOIN by Template -> PayTemplate 这是我目前的尝试:
select
*
FROM
PayTable
WHERE NOT EXISTS (
SELECT 1
FROM
Jobs
LEFT OUTER JOIN PayTemplate
ON PayTemplate.Template = Jobs.CraftTemplate
LEFT OUTER JOIN PayCraftTemplate
ON PayCraftTemplate.Template = PayTemplate.Template
WHERE
PayTable.Craft = PayCraftTemplate.Craft AND PayTable.Job = Jobs.Job
) AND PayTable.Job IS NOT NULL AND PayTable.Craft IS NOT NULL这不是返回我期望的数据,我希望PayTable的第3行只返回1,2行
发布于 2017-02-17 22:00:58
我猜你移动了一些东西来发布这个问题,并基本上解决了它。在上面的查询中,您只需要将PayCraftTemplate.Template更改为PayCraftTemplate.PayTemplate
起初,我认为这是left join的问题,因为where无意中将其转换为inner join,但在这种情况下,无论如何都需要inner join,所以这不是问题所在。
select *
from PayTable
where not exists (
select 1
from Jobs
inner join PayTemplate
on PayTemplate.Template = Jobs.CraftTemplate
and Jobs.Job = PayTable.Job
inner join PayCraftTemplate
on PayCraftTemplate.PayTemplate = PayTemplate.Template
and PayCraftTemplate.Craft = PayTable.Craft
)
and PayTable.Job is not null
and PayTable.Craft is not null结果:
+-------+-----+----------+
| Craft | Job | sequence |
+-------+-----+----------+
| 5501 | 2 | 3 |
+-------+-----+----------+https://stackoverflow.com/questions/42299667
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