Oracle SQL在其具有不同值时获取id

有了这个表，我如何使用select查找超过2个值超过100的ID？

您可以执行以下操作

+----------+---------------------+-------+
| ID       | date_value          | value |
+----------+---------------------+-------+
| 1        | 2011-04-20 21:03:05 | 150   |
| 1        | 2011-04-20 21:03:04 | 150   |
| 2        | 2011-04-20 21:03:03 | 20    |
| 2        | 2011-04-20 21:02:09 | 130   |
| 2        | 2011-04-20 21:02:08 | 130   |
| 3        | 2011-04-20 21:02:07 | 20    |
| ...      | ...                 | ...   |
+----------+---------------------+-------+

SELECT ID, count(*) from mytable
WHERE value > 100
GROUP BY ID
HAVING count(*) >2

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您可以使用以下SQL语句：

select id,count(value)
from table1
where value > 100
having count(value) > 2
group by ID;

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HAVING子句不一定要放在groupby之后吗？它可以放在alsoInteresting之前——这在Oracle中是有效的，但在Postgresql中会起作用。谢谢-我学到了一些东西@psaraj12 having子句不会像您所说的那样工作，因为mlinth saysHI@sudharsan，psaraj是正确的，它在Oracle中工作。看见

SELECT id, count(1)
FROM   mytable
WHERE  value > 100
GROUP BY id
HAVING count(1) > 2;