Sql 选择聚合忽略原因所在

我正在尝试将现有视图转换为可以使用的格式

视图

vw\u temp\u apphoursrasttwoentries

如下所示：

RowNumber | PersNr | Client | Localtion | Agent | Date       | Calweek | Year
----------+--------+--------+-----------+-------+------------+---------+------
1         | 123    | 1      | 1         | ag-01 | 2020-01-01 | 1       | 2021
2         | 123    | 1      | 1         | ag-01 | 2020-01-03 | 1       | 2021
1         | 9999   | 1      | 4         | ag-01 | 2020-01-01 | 1       | 2021
2         | 9999   | 1      | 4         | ag-01 | 2020-01-07 | 1       | 2021

PersNr | Client | Localtion | Agent | minDate    | MaxDate    | DateDiff | Calweek | Year
-------+--------+-----------+-------+------------+------------+----------+---------+-------
123    | 1      | 1         | ag-01 | 2020-01-01 | 2020-01-03 | 3        | 1       | 2021
9999   | 1      | 4         | ag-01 | 2020-01-01 | 2020-01-07 | 7        | 1       | 2021

我需要不同格式的数据，如下所示：

RowNumber | PersNr | Client | Localtion | Agent | Date       | Calweek | Year
----------+--------+--------+-----------+-------+------------+---------+------
1         | 123    | 1      | 1         | ag-01 | 2020-01-01 | 1       | 2021
2         | 123    | 1      | 1         | ag-01 | 2020-01-03 | 1       | 2021
1         | 9999   | 1      | 4         | ag-01 | 2020-01-01 | 1       | 2021
2         | 9999   | 1      | 4         | ag-01 | 2020-01-07 | 1       | 2021

PersNr | Client | Localtion | Agent | minDate    | MaxDate    | DateDiff | Calweek | Year
-------+--------+-----------+-------+------------+------------+----------+---------+-------
123    | 1      | 1         | ag-01 | 2020-01-01 | 2020-01-03 | 3        | 1       | 2021
9999   | 1      | 4         | ag-01 | 2020-01-01 | 2020-01-07 | 7        | 1       | 2021

在原始格式中，一个人只有两行（

行编号1和2

）。我想匹配每一列，并在新视图中具有最小和最大日期以及差异

我的代码：

select a.persnr, a.client, a.location, a.agent, a.calweek, a.year, 
max(a.date) as maxdate, min(b.date) as mindate

    , DATEDIFF(day,a.date,b.date) as dDiff

    from vw_temp_appHoursLastTwoEntries a
       left join vw_temp_appHoursLastTwoEntries b on 
       a.persnr = b.persnr and a.client = b.client and
       a.agent = b.agent and a.date = b.date

    where a.date != b.date and DATEDIFF(day,a.date,b.date) != 0
    or (a.date is not null and b.date is not null) 

    group by a.persnr, a.client, a.location, a.agent, a.calweek, a.year, DATEDIFF(day,a.date,b.date)

问题：

select a.persnr, a.client, a.location, a.agent, a.calweek, a.year, 
max(a.date) as maxdate, min(b.date) as mindate

    , DATEDIFF(day,a.date,b.date) as dDiff

    from vw_temp_appHoursLastTwoEntries a
       left join vw_temp_appHoursLastTwoEntries b on 
       a.persnr = b.persnr and a.client = b.client and
       a.agent = b.agent and a.date = b.date

    where a.date != b.date and DATEDIFF(day,a.date,b.date) != 0
    or (a.date is not null and b.date is not null) 

    group by a.persnr, a.client, a.location, a.agent, a.calweek, a.year, DATEDIFF(day,a.date,b.date)

我现在得到的值似乎是原因不起作用，但我不明白为什么

---

a.date！=b、 date

不应返回

min-

和

maxdates

相同的行。即使

min-

和

maxdates

不同，

datediff

也不会返回任何其他值

0

。

这是您想要的：

declare @Test table (RowNumber int, PersNr int, Client int, Localtion int, Agent varchar(5), [Date] date, Calweek int, [Year] int);

insert into @Test (RowNumber, PersNr, Client, Localtion, Agent, [Date], Calweek, [Year])
values
(1, 123, 1, 1, 'ag-01', '2020-01-01', 1, 2021),
(2, 123, 1, 1, 'ag-01', '2020-01-03', 1, 2021),
(1, 9999, 1, 4, 'ag-01', '2020-01-01', 1, 2021),
(2, 9999, 1, 4, 'ag-01', '2020-01-07', 1, 2021);

select a.PersNr, a.Client, a.Localtion, a.Agent, a.Calweek, a.[Year]
    , max(a.[date]) as maxdate
    , min(b.[date]) as mindate
    , abs(datediff(day,a.[date],b.[date])) as dDiff
from @Test a
left join @Test b on 
    a.persnr = b.persnr and a.client = b.client and
    a.agent = b.agent --and a.[date] = b.[date]
where (/*a.[date] != b.[date] and*/ datediff(day,a.[date],b.[date]) != 0)
  and /*  not OR */ (a.[date] is not null and b.[date] is not null) 
group by a.persnr, a.client, a.Localtion, a.agent, a.calweek, a.[Year], abs(datediff(day,a.[date],b.[date]));

返回：

人事 客户 地点 代理人 卡尔韦克 年 maxdate 心仪的 迪夫 123 1. 1. ag-01 1. 2021 2020-01-03 2020-01-01 2. 9999 1. 4. ag-01 1. 2021 2020-01-07 2020-01-01 6.

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这个条件真的是关于条款的一部分吗？非常感谢！第二次你救了我：）完全错过了，非常感谢你的解释！