Question

给定一个名为Messages的数据库表，其中包含以下列： -

MessageID, FromUser, ToUser, Message, DateTime

我想写一个SQL查询，为“FromUser”列选择不同的值，但我也想要DateTime排序的值。

基本上，查询如：

SELECT * FROM Messages WHERE ToUser='1' ORDER BY DateTime DESC

然后，用以下内容选择不同的FromUser：

SELECT DISTINCT FromUser FROM Messages WHERE ToUser='1'

同时，维护上一次查询的排序。

我尝试使用嵌套查询并遇到问题，因为您无法在内部查询中使用ORDER BY。

基本上，我希望查询的语法表示（这是无效的但是......）：

SELECT DISTINCT FromUser FROM Messages WHERE ToUser=1 AND FromUser IN 
(SELECT * FROM Messages WHERE ToUser=1 ORDER BY DateTime DESC)

Answer 1

我猜这就是你需要的：

;WITH CTE AS
(
    SELECT *, ROW_NUMBER() OVER(PARTITION BY [FromUser] ORDER BY [DateTime] DESC) RN
    FROM [Messages]
    WHERE ToUser='1' 
)
SELECT *
FROM CTE
WHERE RN = 1

Answer 2

因此，请在DateTime语句中加入select，或者使用MAX()进行分组，或者使用其中一个聚合函数，例如Select fromuser, datetime from messages where touser = '1' group by fromuser, datetime order by fromuser, datetime，如：

Select fromuser, max(datetime) as max_datetime
from messages
where touser = '1'
group by 
fromuser

或

{{1}}

Answer 3

您将为每个FromUser设置多个时间戳。我将假设您要为每个时间戳使用最后一个时间戳。

select * 
from Messages join 
    (select fromuser, MAX(datetime) as lastdate from Messages
    group by FromUser) as abb1 on abb1.FromUser = messages.FromUser and abb1.lastdate = messages.Datetime
where ToUser = '1'
order by Datetime desc

Answer 4

SELECT DISTINCT [Messages].[FromUser] FROM (
SELECT [FromUser] as [FromUser], MAX(DateTime) as [Date]
FROM [Messages]
GROUP BY [FromUser]) as [Messages]

Answer 5

另一种方式;

select * from
(
    select *, max(DateTime) over (partition by FromUser) newest
    from Messages
) T
where DateTime = newest

（如果同一个人的日期相同，则为n行）

使用DISTINCT和ORDER BY时出现SQL查询问题

5 个答案: