Question

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

我正在尝试查询工作，显示每个月的总计。到目前为止，我已经尝试过DateDiff和嵌套选择，但两者都没有给我我想要的东西。这是我认为最接近的：

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

但是当我得到值时，我只是得到空值。我错过了什么吗？

Answer 1

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

Answer 2

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]

Answer 3

我更喜欢合并DATEADD和DATEDIFF这样的函数：

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

这两个函数一起将日期组件 less 归零而不是指定的 datepart （在本例中为MONTH）。

您可以将datepart位更改为YEAR，WEEK，DAY等...这非常方便。

您的原始SQL查询将看起来像这样（我无法测试它，因为我没有您的数据集，但它应该让您走上正确的轨道）。

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列被输入为DateTime，如果您需要进一步处理该数据或将其映射到.NET对象，这也是一个很好的优势。

Answer 4

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

应该......

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

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

初始化后尝试SELECT @start以检查这是否正确。

Answer 5

如果您需要经常这样做，我可能会在表格中添加一个计算列PaymentMonth：

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

它持久化并存储在表中 - 因此查询它确实没有性能开销。它是一个4字节的INT值 - 因此空间开销也很小。

一旦你有了这个，你可以简化你的查询，使其成为：

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

Answer 6

另一种不涉及在结果中添加列的方法是简单地将日期的day组件清空，因此2016-07-13和2016-07-16都会是2016-07-01 - 从而使它们按月平等。

如果您有date（不是datetime）值，则可以直接将其归零：

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个值，则需要使用CONVERT删除时间部分：

SELECT
    DATEADD( day, 1 - DATEPART( day, [Date] ),  CONVERT( date, [Date] ) ),
    COUNT(*)
FROM
    [Table]
GROUP BY
    DATEADD( day, 1 - DATEPART( day, [Date] ),  CONVERT( date, [Date] ) )

Answer 7

现在您的查询只是明确查看年份= 2010年的付款，但是，我认为您的意思是让您的Jan / Feb / Mar实际代表2009年。如果是这样，您需要针对该情况稍微调整一下。不要继续重新查询每列的总和值，只是以月为单位的日期差异的条件。将其余部分放在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

SQL Server Group By Month

7 个答案: