Question

我需要通过RelationShipId选择3个最新的唯一行（因此，没有重复的行），其中一个时间戳记与今天最接近。

我现在有一个看起来像这样的查询，但是它不起作用。

我得到一个：：在选择列表中无效，因为它没有包含在聚合函数或GROUP BY子句中。

到目前为止我拥有的脚本：

SELECT * FROM PlayerAction
WHERE [ActionType] = 'SomeString' AND [RelationShipType] = 'SomeString' AND [PlayerId] = 5 AND [Deleted] = 0
GROUP BY RelationShipId
ORDER BY [TimeStamp] DESC
OFFSET 1 ROWS
FETCH NEXT 3 ROWS ONLY

SampleData

{
 PlayerId=5
 ActionType="SomeString"
 RelationShipType="SomeString"
 RelationShipId=1
 TimeStamp=2019-07-04 07:45:47.400
 Deleted=0
}
{
 PlayerId=5
 ActionType="SomeString"
 RelationShipType="SomeString"
 RelationShipId=2
 TimeStamp=2019-07-03 14:29:10.530
 Deleted=0
}
{
 PlayerId=5
 ActionType="SomeString"
 RelationShipType="SomeString"
 RelationShipId=3
 TimeStamp=2019-07-03 13:56:26.057
 Deleted=0
}

想象有5行具有相同的RelationShipId，无休止的RelationShipIds。我想要时间戳记的前3个唯一行，最后三个PlayerActions

Answer 1

看来您需要与众不同的

SELECT distinct top 3 * FROM PlayerAction
        WHERE [ActionType] = 'SomeString' AND [RelationShipType] = 'SomeString' AND [PlayerId] = SomeNumber AND [Deleted] = 0
       ORDER BY [TimeStamp] DESC

选择列中具有唯一值的行（按时间戳记排序）？

1 个答案: