SQL Server按月分组

我有一个具有此架构的表

ItemID UserID Year IsPaid PaymentDate Amount

1 1 2009 0 2009-11-01 300

2 1 2009 0 2009-12-01 342

3 1 2010 0 2010-01-01 243

4 1 2010 0 2010-02-01 2543

5 1 2010 0 2010-03-01 475

我正在尝试查询显示每个月的总数的查询。 到目前为止，我已经尝试过DateDiff和嵌套选择，但都没有给我想要的东西。 我认为这是最接近的：

DECLARE @start [datetime] = 2010/4/1;

SELECT ItemID, IsPaid,

(SELECT SUM(Amount) FROM Payments WHERE Year = 2010 And DateDiff(m, PaymentDate, @start) = 0 AND UserID = 100) AS "Apr",

(SELECT SUM(Amount) FROM Payments WHERE Year = 2010 AND DateDiff(m, PaymentDate, @start) =1 AND UserID = 100) AS "May",

(SELECT SUM(Amount) FROM Payments WHERE Year = 2010 AND DateDiff(m, PaymentDate, @start) =2 AND UserID = 100) AS "Jun",

(SELECT SUM(Amount) FROM Payments WHERE Year = 2010 AND DateDiff(m, PaymentDate, @start) =3 AND UserID = 100) AS "Jul",

(SELECT SUM(Amount) FROM Payments WHERE Year = 2010 AND DateDiff(m, PaymentDate, @start) =4 AND UserID = 100) AS "Aug",

(SELECT SUM(Amount) FROM Payments WHERE Year = 2010 AND DateDiff(m, PaymentDate, @start) =5 AND UserID = 100) AS "Sep",

(SELECT SUM(Amount) FROM Payments WHERE Year = 2010 AND DateDiff(m, PaymentDate, @start) =6 AND UserID = 100) AS "Oct",

(SELECT SUM(Amount) FROM Payments WHERE Year = 2010 AND DateDiff(m, PaymentDate, @start) =7 AND UserID = 100) AS "Nov",

(SELECT SUM(Amount) FROM Payments WHERE Year = 2010 AND DateDiff(m, PaymentDate, @start) =8 AND UserID = 100) AS "Dec",

(SELECT SUM(Amount) FROM Payments WHERE Year = 2010 AND DateDiff(m, PaymentDate, @start) =9 AND UserID = 100) AS "Jan",

(SELECT SUM(Amount) FROM Payments WHERE Year = 2010 AND DateDiff(m, PaymentDate, @start) =10 AND UserID = 100) AS "Feb",

(SELECT SUM(Amount) FROM Payments WHERE Year = 2010 AND DateDiff(m, PaymentDate, @start) =11 AND UserID = 100) AS "Mar"

FROM LIVE L INNER JOIN Payments I ON I.LiveID = L.RECORD_KEY

WHERE UserID = 16178

但是当我应该获取值时，我只会得到空值。 我想念什么吗？

7个解决方案

101 votes

SELECT CONVERT(NVARCHAR(10), PaymentDate, 120) [Month], SUM(Amount) [TotalAmount]

FROM Payments

GROUP BY CONVERT(NVARCHAR(10), PaymentDate, 120)

ORDER BY [Month]

您也可以尝试：

SELECT DATEPART(Year, PaymentDate) Year, DATEPART(Month, PaymentDate) Month, SUM(Amount) [TotalAmount]

FROM Payments

GROUP BY DATEPART(Year, PaymentDate), DATEPART(Month, PaymentDate)

ORDER BY Year, Month

Dave Downs answered 2020-06-17T06:16:37Z

20 votes

将NVARCHAR的尺寸限制为7，提供给CONVERT以仅显示“ YYYY-MM”

SELECT CONVERT(NVARCHAR(7),PaymentDate,120) [Month], SUM(Amount) [TotalAmount]

FROM Payments

GROUP BY CONVERT(NVARCHAR(7),PaymentDate,120)

ORDER BY [Month]

Martyn Davis answered 2020-06-17T06:16:58Z

5 votes

我更喜欢结合Month和DateTime函数，如下所示：

GROUP BY DATEADD(MONTH, DATEDIFF(MONTH, 0, Created),0)

这两个函数共同将比指定日期部分(在此示例中为Month)小的日期分量归零。

您可以将datepart位更改为Month、DateTime、DAY等，这非常方便。

这样，您原始的SQL查询将类似于以下内容(由于我没有设置数据，因此无法对其进行测试，但它应该使您处于正确的轨道)。

DECLARE @start [datetime] = '2010-04-01';

SELECT

ItemID,

UserID,

DATEADD(MONTH, DATEDIFF(MONTH, 0, Created),0) [Month],

IsPaid,

SUM(Amount)

FROM LIVE L

INNER JOIN Payments I ON I.LiveID = L.RECORD_KEY

WHERE UserID = 16178

AND PaymentDate > @start

还有一件事：Month列键入为DateTime，如果您需要进一步处理该数据或将其映射为.NET对象，这也是一个不错的优势。

bounav answered 2020-06-17T06:17:36Z

3 votes

DECLARE @start [datetime] = 2010/4/1;

应该...

DECLARE @start [datetime] = '2010-04-01';

您所拥有的是将2010除以4，然后除以1，然后转换为日期。 从1900-01-01开始的第57.5天。

初始化后，尝试SELECT @start，以检查是否正确。

MatBailie answered 2020-06-17T06:18:05Z

3 votes

如果您需要经常执行此操作，则可能要向表中添加一个计算列PaymentMonth：

ALTER TABLE dbo.Payments ADD PaymentMonth AS MONTH(PaymentDate) PERSISTED

它被持久化并存储在表中-因此查询它实际上没有性能开销。 这是一个4字节的INT值-因此空间开销也很小。

一旦有了它，就可以简化查询，使其类似于以下内容：

SELECT ItemID, IsPaid,

(SELECT SUM(Amount) FROM Payments WHERE Year = 2010 And PaymentMonth = 1 AND UserID = 100) AS 'Jan',

(SELECT SUM(Amount) FROM Payments WHERE Year = 2010 And PaymentMonth = 2 AND UserID = 100) AS 'Feb',

.... and so on .....

FROM LIVE L

INNER JOIN Payments I ON I.LiveID = L.RECORD_KEY

WHERE UserID = 16178

marc_s answered 2020-06-17T06:18:34Z

1 votes

另一种不涉及在结果中添加列的方法是简单地将日期的CONVERT零置零，因此CONVERT和2016-07-16都将是2016-07-01-从而使它们按月相等。

如果您具有datetime(而不是CONVERT)值，则可以将其直接置零：

SELECT

DATEADD( day, 1 - DATEPART( day, [Date] ), [Date] ),

COUNT(*)

FROM

[Table]

GROUP BY

DATEADD( day, 1 - DATEPART( day, [Date] ), [Date] )

如果您有datetime值，则需要使用CONVERT删除“时段”部分：

SELECT

DATEADD( day, 1 - DATEPART( day, [Date] ), CONVERT( date, [Date] ) ),

COUNT(*)

FROM

[Table]

GROUP BY

DATEADD( day, 1 - DATEPART( day, [Date] ), CONVERT( date, [Date] ) )

Dai answered 2020-06-17T06:19:03Z

0 votes

现在，您的查询明确地只查看了年份= 2010的付款，但是，我想您的意思是让Jan / Feb / Mar实际代表2009。如果是这样，则需要针对这种情况进行一些调整。 不要继续查询每一列的总和值，而只是查询以月为单位的日期差的条件。 将其余的放在WHERE子句中。

SELECT

SUM( case when DateDiff(m, PaymentDate, @start) = 0

then Amount else 0 end ) AS "Apr",

SUM( case when DateDiff(m, PaymentDate, @start) = 1

then Amount else 0 end ) AS "May",

SUM( case when DateDiff(m, PaymentDate, @start) = 2

then Amount else 0 end ) AS "June",

SUM( case when DateDiff(m, PaymentDate, @start) = 3

then Amount else 0 end ) AS "July",

SUM( case when DateDiff(m, PaymentDate, @start) = 4

then Amount else 0 end ) AS "Aug",

SUM( case when DateDiff(m, PaymentDate, @start) = 5

then Amount else 0 end ) AS "Sep",

SUM( case when DateDiff(m, PaymentDate, @start) = 6

then Amount else 0 end ) AS "Oct",

SUM( case when DateDiff(m, PaymentDate, @start) = 7

then Amount else 0 end ) AS "Nov",

SUM( case when DateDiff(m, PaymentDate, @start) = 8

then Amount else 0 end ) AS "Dec",

SUM( case when DateDiff(m, PaymentDate, @start) = 9

then Amount else 0 end ) AS "Jan",

SUM( case when DateDiff(m, PaymentDate, @start) = 10

then Amount else 0 end ) AS "Feb",

SUM( case when DateDiff(m, PaymentDate, @start) = 11

then Amount else 0 end ) AS "Mar"

FROM

Payments I

JOIN Live L

on I.LiveID = L.Record_Key

WHERE

Year = 2010

AND UserID = 100

DRapp answered 2020-06-17T06:19:24Z