[英]MSSQL query: how to find the incorrect row by each partition?
我需要根据逻辑找到不正确的行。
逻辑是:
如果孩子有排(我会打电话给第一排)
| merit | fruit | vegetable | | --------- | ----- | --------- | | behaviour | apple | cucumber |
然后在有优点的行=诗和水果=苹果必须只有蔬菜= cucumber (黄瓜没有别的词) (这是第二行)
| merit | fruit | vegetable | | ----- | ----- | --------- | | poem | apple | cucumber |
第二行的 AND 时间间隔必须比第一行的时间早或晚 4 小时,作为正确示例:
| child_id | date | merit | fruit | vegetable | | --------- | --------------- | --------- | ----- | --------- | | 2 | 1/26/2022 16:00 | poem | apple | cucumber | | 2 | 1/26/2022 18:00 | behaviour | apple | cucumber |
如我们所见,间隔为 4 小时
我有桌子:
| child_id | date | merit | fruit | vegetable |
| --------- | --------------- | ----------- | ------- | --------- |
| 1 | 1/27/2022 14:00 | behaviour | apple | cucumber |
| 1 | 1/27/2022 15:00 | poem | apple | carrot |
| 1 | 1/27/2022 17:00 | sleep | apple | ginger |
| 1 | 1/27/2022 20:00 | competition | berry | tomatoe |
| 2 | 1/26/2022 13:00 | sleep | apricot | tomatoe |
| 2 | 1/30/2022 13:00 | poem | apple | cucumber |
| 2 | 1/29/2022 13:00 | poem | apple | cucumber |
| 2 | 1/26/2022 16:00 | poem | apple | cucumber |
| 2 | 1/26/2022 18:00 | behaviour | apple | cucumber |
| 2 | 1/26/2022 19:00 | present | apple | broccoli |
| 3 | 1/25/2022 11:00 | present | orange | cucumber |
| 3 | 1/25/2022 13:00 | poem | apple | ginger |
| 3 | 1/25/2022 15:00 | behaviour | apple | cucumber |
| 4 | 1/26/2022 14:00 | behaviour | apple | cucumber |
| 4 | 1/27/2022 21:00 | poem | apple | carrot |
| 4 | 1/27/2022 15:00 | poem | apple | carrot |
| 4 | 1/27/2022 20:00 | sleep | apple | ginger |
| 4 | 1/27/2022 21:00 | competition | berry | tomatoe |
我期望的结果:
| child_id | date | merit | fruit | vegetable |
| --------- | --------------- | ----- | ----- | --------- |
| 1 | 1/27/2022 15:00 | poem | apple | carrot |
| 3 | 1/25/2022 13:00 | poem | apple | ginger |
我不知道如何按孩子找到这些行。 我写了这个 SQL 并卡住了:
select * from example_1 where merit in ('behaviour', 'poem')
我这里需要分区吗?
在这种方法中,我们使用整理子查询。 顶部查询 B 为所需结果定义数据的非连接限制。 所以菜<>cucumber和功=诗
Exists 确保定义了第一行的限制并且存在非匹配项的相关性。 所以我们确保水果匹配,优点是“行为”,child_id 的匹配,并且日期的差异在 4 小时内。
SELECT B.*
FROM table B
WHERE vegetable <> 'cucumber'
and merit = 'poem'
and exists (SELECT 1
FROM Table A
WHERE A.Fruit = B.Fruit
AND A.Child_id = B.Child_ID
AND A.merit = 'behaviour'
AND abs(Datediff(hour,A.Date,B.Date)) <=4)
给我们:
+----------+-------------------------+-------+-------+-----------+
| child_id | date | merit | fruit | vegetable |
+----------+-------------------------+-------+-------+-----------+
| 1 | 2022-01-27 15:00:00.000 | poem | apple | carrot |
| 3 | 2022-01-25 13:00:00.000 | poem | apple | ginger |
+----------+-------------------------+-------+-------+-----------+
一种可能的解决方案是使用 LEFT OUTER JOIN 将表连接到自身,然后只接受表的连接版本返回 null 的记录:
SELECT e1.*
FROM example_1 e1
LEFT OUTER JOIN example_1 e2
ON e1.fruit = e2.fruit
AND e1.vegetable <> e2.vegetable
AND e2.date BETWEEN DATEADD(HOUR, -4, e1.date) AND e1.date
AND e2.merit = 'behavior'
WHERE e1.merit = 'poem'
AND e2.child_id IS NULL
诀窍主要在连接标准中,我们希望确保我们在“行为”和“诗歌”之间匹配vegetable
,同时还要检查最后 4 小时。
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