SQL Server-一周结束

我有下面的T-SQL代码可以工作，但我认为可能会减少，但不确定如何减少

SELECT 
    datepart(YEAR, CONVERT(DATE, GETDATE())) AS 'Year', 
    Cast(Datepart(month, DATEADD (D, -1 * DatePart (DW, getdate()) + 7, getdate())) AS VARCHAR) + '-' + Cast(Datepart(DAY, DATEADD (D, -1 * DatePart (DW, getdate()) + 7, getdate()))AS VARCHAR) [Week Ending]

我只想在另一列中以年为结束的一周，这样我就可以按年分组，然后按周结束。我能帮忙吗

你不能这样做吗？：

DATEADD(DAY, 7 - DATEPART(WEEKDAY, GETDATE()), CAST(GETDATE() AS DATE)) [WeekEnd]

没有必要在单独的栏中单独列出年份

编辑：我不知道你的评论是什么意思。这会打印出日期，并且是日期格式，而不是varchar，而不是周数

示例：

DECLARE @dateTime datetime
SET     @dateTime = '2016-01-01'
select  DATEADD(DAY, 7 - DATEPART(WEEKDAY, @dateTime), CAST(@dateTime AS DATE)) [WeekEnd]

2016-01-02

2016-01-09

使用示例：

DECLARE @table TABLE (orderDate datetime, orderAmount float)
INSERT INTO @table
(
    orderDate,
    orderAmount
)
SELECT '2015-01-02', 500
union all SELECT '2015-01-04', 500
union all SELECT '2015-01-05', 500
union all SELECT '2015-01-05', 500
union all SELECT '2015-01-06', 500
union ALL SELECT '2015-01-11', 400
union all SELECT '2016-01-01', 500
union all SELECT '2016-01-02', 500
union all SELECT '2016-01-04', 500
union all SELECT '2016-01-05', 500
union all SELECT '2016-01-05', 500
union all SELECT '2016-01-06', 500
UNION ALL SELECT '2016-01-11', 400
UNION ALL SELECT '2016-12-11', 1200

每个日期的订单金额：

SELECT  
    orderDate, 
    sum(orderAmount) AS orderSumForGrouping, 
    count(1) AS numberOfOrdersWithinGrouping
FROM    @table o
GROUP   BY orderDate

输出：

orderDate   orderSumForGrouping numberOfOrdersWithinGrouping
2015-01-02 00:00:00.000 500 1
2015-01-04 00:00:00.000 500 1
2015-01-05 00:00:00.000 1000    2
2015-01-06 00:00:00.000 500 1
2015-01-11 00:00:00.000 400 1
2016-01-01 00:00:00.000 500 1
2016-01-02 00:00:00.000 500 1
2016-01-04 00:00:00.000 500 1
2016-01-05 00:00:00.000 1000    2
2016-01-06 00:00:00.000 500 1
2016-01-11 00:00:00.000 400 1
2016-12-11 00:00:00.000 1200    1

orderYear   orderSumForGrouping numberOfOrdersWithinGrouping
2015    2900    6
2016    4600    8

ordersPerWeek   orderSumForGrouping numberOfOrdersWithinGrouping
2015-01-03  500 1
2015-01-10  2000    4
2015-01-17  400 1
2016-01-02  1000    2
2016-01-09  2000    4
2016-01-16  400 1
2016-12-17  1200    1

按年份分组的订单金额：

SELECT  
    year(orderDate) AS orderYear, 
    sum(orderAmount) AS orderSumForGrouping, 
    count(1) AS numberOfOrdersWithinGrouping
FROM    @table o
GROUP   BY year(orderDate)

输出：

orderDate   orderSumForGrouping numberOfOrdersWithinGrouping
2015-01-02 00:00:00.000 500 1
2015-01-04 00:00:00.000 500 1
2015-01-05 00:00:00.000 1000    2
2015-01-06 00:00:00.000 500 1
2015-01-11 00:00:00.000 400 1
2016-01-01 00:00:00.000 500 1
2016-01-02 00:00:00.000 500 1
2016-01-04 00:00:00.000 500 1
2016-01-05 00:00:00.000 1000    2
2016-01-06 00:00:00.000 500 1
2016-01-11 00:00:00.000 400 1
2016-12-11 00:00:00.000 1200    1

orderYear   orderSumForGrouping numberOfOrdersWithinGrouping
2015    2900    6
2016    4600    8

ordersPerWeek   orderSumForGrouping numberOfOrdersWithinGrouping
2015-01-03  500 1
2015-01-10  2000    4
2015-01-17  400 1
2016-01-02  1000    2
2016-01-09  2000    4
2016-01-16  400 1
2016-12-17  1200    1

按周末日期分组的订单金额：

SELECT  
    DATEADD(DAY, 7 - DATEPART(WEEKDAY, orderDate), CAST(orderDate AS DATE)) AS ordersPerWeek, 
    sum(orderAmount) AS orderSumForGrouping, 
    count(1) AS numberOfOrdersWithinGrouping
FROM    @table o
GROUP   BY DATEADD(DAY, 7 - DATEPART(WEEKDAY, orderDate), CAST(orderDate AS DATE))

输出：

orderDate   orderSumForGrouping numberOfOrdersWithinGrouping
2015-01-02 00:00:00.000 500 1
2015-01-04 00:00:00.000 500 1
2015-01-05 00:00:00.000 1000    2
2015-01-06 00:00:00.000 500 1
2015-01-11 00:00:00.000 400 1
2016-01-01 00:00:00.000 500 1
2016-01-02 00:00:00.000 500 1
2016-01-04 00:00:00.000 500 1
2016-01-05 00:00:00.000 1000    2
2016-01-06 00:00:00.000 500 1
2016-01-11 00:00:00.000 400 1
2016-12-11 00:00:00.000 1200    1

orderYear   orderSumForGrouping numberOfOrdersWithinGrouping
2015    2900    6
2016    4600    8

ordersPerWeek   orderSumForGrouping numberOfOrdersWithinGrouping
2015-01-03  500 1
2015-01-10  2000    4
2015-01-17  400 1
2016-01-02  1000    2
2016-01-09  2000    4
2016-01-16  400 1
2016-12-17  1200    1

请注意，

2016-01-02

、

2016-01-09

和

2016-01-16

都是星期六-一周的最后一天。（其他日期也适用，但这些日期刚刚出现，因此最容易检查）

您不能这样做吗？：

DATEADD(DAY, 7 - DATEPART(WEEKDAY, GETDATE()), CAST(GETDATE() AS DATE)) [WeekEnd]

没有必要在单独的栏中单独列出年份

编辑：我不知道你的评论是什么意思。这会打印出日期，并且是日期格式，而不是varchar，而不是周数

示例：

DECLARE @dateTime datetime
SET     @dateTime = '2016-01-01'
select  DATEADD(DAY, 7 - DATEPART(WEEKDAY, @dateTime), CAST(@dateTime AS DATE)) [WeekEnd]

2016-01-02

2016-01-09

使用示例：

DECLARE @table TABLE (orderDate datetime, orderAmount float)
INSERT INTO @table
(
    orderDate,
    orderAmount
)
SELECT '2015-01-02', 500
union all SELECT '2015-01-04', 500
union all SELECT '2015-01-05', 500
union all SELECT '2015-01-05', 500
union all SELECT '2015-01-06', 500
union ALL SELECT '2015-01-11', 400
union all SELECT '2016-01-01', 500
union all SELECT '2016-01-02', 500
union all SELECT '2016-01-04', 500
union all SELECT '2016-01-05', 500
union all SELECT '2016-01-05', 500
union all SELECT '2016-01-06', 500
UNION ALL SELECT '2016-01-11', 400
UNION ALL SELECT '2016-12-11', 1200

每个日期的订单金额：

SELECT  
    orderDate, 
    sum(orderAmount) AS orderSumForGrouping, 
    count(1) AS numberOfOrdersWithinGrouping
FROM    @table o
GROUP   BY orderDate

输出：

orderDate   orderSumForGrouping numberOfOrdersWithinGrouping
2015-01-02 00:00:00.000 500 1
2015-01-04 00:00:00.000 500 1
2015-01-05 00:00:00.000 1000    2
2015-01-06 00:00:00.000 500 1
2015-01-11 00:00:00.000 400 1
2016-01-01 00:00:00.000 500 1
2016-01-02 00:00:00.000 500 1
2016-01-04 00:00:00.000 500 1
2016-01-05 00:00:00.000 1000    2
2016-01-06 00:00:00.000 500 1
2016-01-11 00:00:00.000 400 1
2016-12-11 00:00:00.000 1200    1

orderYear   orderSumForGrouping numberOfOrdersWithinGrouping
2015    2900    6
2016    4600    8

ordersPerWeek   orderSumForGrouping numberOfOrdersWithinGrouping
2015-01-03  500 1
2015-01-10  2000    4
2015-01-17  400 1
2016-01-02  1000    2
2016-01-09  2000    4
2016-01-16  400 1
2016-12-17  1200    1

按年份分组的订单金额：

SELECT  
    year(orderDate) AS orderYear, 
    sum(orderAmount) AS orderSumForGrouping, 
    count(1) AS numberOfOrdersWithinGrouping
FROM    @table o
GROUP   BY year(orderDate)

输出：

orderDate   orderSumForGrouping numberOfOrdersWithinGrouping
2015-01-02 00:00:00.000 500 1
2015-01-04 00:00:00.000 500 1
2015-01-05 00:00:00.000 1000    2
2015-01-06 00:00:00.000 500 1
2015-01-11 00:00:00.000 400 1
2016-01-01 00:00:00.000 500 1
2016-01-02 00:00:00.000 500 1
2016-01-04 00:00:00.000 500 1
2016-01-05 00:00:00.000 1000    2
2016-01-06 00:00:00.000 500 1
2016-01-11 00:00:00.000 400 1
2016-12-11 00:00:00.000 1200    1

orderYear   orderSumForGrouping numberOfOrdersWithinGrouping
2015    2900    6
2016    4600    8

ordersPerWeek   orderSumForGrouping numberOfOrdersWithinGrouping
2015-01-03  500 1
2015-01-10  2000    4
2015-01-17  400 1
2016-01-02  1000    2
2016-01-09  2000    4
2016-01-16  400 1
2016-12-17  1200    1

按周末日期分组的订单金额：

SELECT  
    DATEADD(DAY, 7 - DATEPART(WEEKDAY, orderDate), CAST(orderDate AS DATE)) AS ordersPerWeek, 
    sum(orderAmount) AS orderSumForGrouping, 
    count(1) AS numberOfOrdersWithinGrouping
FROM    @table o
GROUP   BY DATEADD(DAY, 7 - DATEPART(WEEKDAY, orderDate), CAST(orderDate AS DATE))

输出：

orderDate   orderSumForGrouping numberOfOrdersWithinGrouping
2015-01-02 00:00:00.000 500 1
2015-01-04 00:00:00.000 500 1
2015-01-05 00:00:00.000 1000    2
2015-01-06 00:00:00.000 500 1
2015-01-11 00:00:00.000 400 1
2016-01-01 00:00:00.000 500 1
2016-01-02 00:00:00.000 500 1
2016-01-04 00:00:00.000 500 1
2016-01-05 00:00:00.000 1000    2
2016-01-06 00:00:00.000 500 1
2016-01-11 00:00:00.000 400 1
2016-12-11 00:00:00.000 1200    1

orderYear   orderSumForGrouping numberOfOrdersWithinGrouping
2015    2900    6
2016    4600    8

ordersPerWeek   orderSumForGrouping numberOfOrdersWithinGrouping
2015-01-03  500 1
2015-01-10  2000    4
2015-01-17  400 1
2016-01-02  1000    2
2016-01-09  2000    4
2016-01-16  400 1
2016-12-17  1200    1

---

请注意，

2016-01-02

、

2016-01-09

和

2016-01-16

都是星期六-一周的最后一天。（其他日期也适用，但这些日期刚刚出现，因此最容易检查）

我可以将其缩小一点的一种方法是为

WeekEnd

创建一个自动计算列，并在查询中重用它：

设置：

-- drop table testTable
create table testTable
(
    MyDate SMALLDATETIME,
    WeekEnd AS DATEADD (D, -1 * DatePart(DW, MyDate) + 7, MyDate)
)
GO

INSERT INTO testTable VALUES ('20141201'), ('20150101'), ('20150130')
GO

select * from testTable
GO

查询：

select CAST(Datepart(month, WeekEnd) AS VARCHAR) + ' - ' + CAST(Datepart(day, WeekEnd) AS VARCHAR) AS [Week Ending]
from testTable

---

不幸的是，我无法使列持久化（我无法说服SQL Server我所做的是确定性的，即使使用带有模式绑定的函数）。这将非常好，因为持久化的计算列可以被索引

我可以将其缩小一点的一种方法是为

WeekEnd

创建一个自动计算列，并在查询中重用它：

设置：

-- drop table testTable
create table testTable
(
    MyDate SMALLDATETIME,
    WeekEnd AS DATEADD (D, -1 * DatePart(DW, MyDate) + 7, MyDate)
)
GO

INSERT INTO testTable VALUES ('20141201'), ('20150101'), ('20150130')
GO

select * from testTable
GO

查询：

select CAST(Datepart(month, WeekEnd) AS VARCHAR) + ' - ' + CAST(Datepart(day, WeekEnd) AS VARCHAR) AS [Week Ending]
from testTable

---

不幸的是，我无法使列持久化（我无法说服SQL Server我所做的是确定性的，即使使用带有模式绑定的函数）。这将非常好，因为持久化的计算列可以被索引

这就是我所做的，但我想要的是结束一周的实际日期，而不是周数。我发布的内容给出的是实际日期，而不是周数。我想我是这样开始的，但这肯定更好。有时你走错了路，需要从头开始。这就是我所做的，但我想要的是结束一周的实际日期，而不是周数。我发布的内容给出的是实际日期，而不是周数。我想我是这样开始的，但确实更好。有时你走错了路，需要从头开始。对于年份，你可以使用内置函数

Year

像

Year（GETDATE（））一样作为“Year”

，而不是强迫我对代码进行反向工程，并对你的语言设置进行假设，你如何定义一周结束？周末是星期六、星期天还是别的什么？而且，即使基于此，您希望计算什么值？对于年，您可以使用内置函数

Year

像

Year（GETDATE（））一样作为“年”

，而不是强迫我对代码进行反向工程，并对您的语言设置进行假设，您如何定义一周结束？周末是星期六、星期天还是别的什么？即使基于此，你想计算什么值？