SQL阶梯查询
SQL阶梯查询
提问于 2012-06-27 05:02:48
我需要一些帮助，以产生一个MS SQL 2012查询，将匹配所需的阶梯输出。行按一个日期范围(帐户提交日期月)汇总数据，列按另一个日期范围(付款日期月)汇总数据
表1： accounts 跟踪为收集而放置的帐户。
CREATE TABLE [dbo].[Accounts](
[AccountID] [nchar](10) NOT NULL,
[SubmissionDate] [date] NOT NULL,
[Amount] [money] NOT NULL,
CONSTRAINT [PK_Accounts] PRIMARY KEY CLUSTERED (AccountID ASC))
INSERT INTO [dbo].[Accounts] VALUES ('1000', '2012-01-01', 1999.00)
INSERT INTO [dbo].[Accounts] VALUES ('1001', '2012-01-02', 100.00)
INSERT INTO [dbo].[Accounts] VALUES ('1002', '2012-02-05', 350.00)
INSERT INTO [dbo].[Accounts] VALUES ('1003', '2012-03-01', 625.00)
INSERT INTO [dbo].[Accounts] VALUES ('1004', '2012-03-10', 50.00)
INSERT INTO [dbo].[Accounts] VALUES ('1005', '2012-03-10', 10.00)表2：Trans跟踪付款
CREATE TABLE [dbo].[Trans](
[TranID] [int] IDENTITY(1,1) NOT NULL,
[AccountID] [nchar](10) NOT NULL,
[TranDate] [date] NOT NULL,
[TranAmount] [money] NOT NULL,
CONSTRAINT [PK_Trans] PRIMARY KEY CLUSTERED (TranID ASC))
INSERT INTO [dbo].[Trans] VALUES (1000, '2012-01-15', 300.00)
INSERT INTO [dbo].[Trans] VALUES (1000, '2012-02-15', 300.00)
INSERT INTO [dbo].[Trans] VALUES (1000, '2012-03-15', 300.00)
INSERT INTO [dbo].[Trans] VALUES (1002, '2012-02-20', 325.00)
INSERT INTO [dbo].[Trans] VALUES (1002, '2012-04-20', 25.00)
INSERT INTO [dbo].[Trans] VALUES (1003, '2012-03-24', 625.00)
INSERT INTO [dbo].[Trans] VALUES (1004, '2012-03-28', 31.00)
INSERT INTO [dbo].[Trans] VALUES (1004, '2012-04-12', 5.00)
INSERT INTO [dbo].[Trans] VALUES (1005, '2012-04-08', 7.00)
INSERT INTO [dbo].[Trans] VALUES (1005, '2012-04-28', 3.00)下面是期望的输出应该是什么样子
*Total Payments in Each Month*
SubmissionYearMonth TotalAmount | 2012-01 2012-02 2012-03 2012-04
--------------------------------------------------------------------
2012-01 2099.00 | 300.00 300.00 300.00 0.00
2012-02 350.00 | 325.00 0.00 25.00
2012-03 685.00 | 656.00 15.00前两列按月份对Account.Amount分组求和。
最后4列按月对当前行的给定月份中的帐户的Tran.TranAmount求和。
我一直在处理的查询感觉很接近。我只是不知道延迟是否正确。到目前为止，我正在使用的查询如下：
Select SubmissionYearMonth,
TotalAmount,
pt.[0] AS MonthOld0,
pt.[1] AS MonthOld1,
pt.[2] AS MonthOld2,
pt.[3] AS MonthOld3,
pt.[4] AS MonthOld4,
pt.[5] AS MonthOld5,
pt.[6] AS MonthOld6,
pt.[7] AS MonthOld7,
pt.[8] AS MonthOld8,
pt.[9] AS MonthOld9,
pt.[10] AS MonthOld10,
pt.[11] AS MonthOld11,
pt.[12] AS MonthOld12,
pt.[13] AS MonthOld13
From (
SELECT Convert(Char(4),Year(SubmissionDate)) + '-' + Right('00' + Convert(VarChar(2), DatePart(Month, SubmissionDate)),2) AS SubmissionYearMonth,
SUM(Amount) AS TotalAmount
FROM Accounts
GROUP BY Convert(Char(4),Year(SubmissionDate)) + '-' + Right('00' + Convert(VarChar(2), DatePart(Month, SubmissionDate)),2)
)
AS AccountSummary
OUTER APPLY
(
SELECT *
FROM (
SELECT CASE WHEN DATEDIFF(Month, SubmissionDate, TranDate) < 13
THEN DATEDIFF(Month, SubmissionDate, TranDate)
ELSE 13
END AS PaymentMonthAge,
TranAmount
FROM Trans INNER JOIN Accounts ON Trans.AccountID = Accounts.AccountID
Where Convert(Char(4),Year(TranDate)) + '-' + Right('00' + Convert(VarChar(2), DatePart(Month, TranDate)),2)
= AccountSummary.SubmissionYearMonth
) as TransTemp
PIVOT (SUM(TranAmount)
FOR PaymentMonthAge IN ([0],
[1],
[2],
[3],
[4],
[5],
[6],
[7],
[8],
[9],
[10],
[11],
[12],
[13])) as TransPivot
) as pt它会产生以下输出：
SubmissionYearMonth TotalAmount MonthOld0 MonthOld1 MonthOld2 MonthOld3 ...
2012-01 2099.00 300.00 NULL NULL NULL ...
2012-02 350.00 325.00 300.00 NULL NULL ...
2012-03 685.00 656.00 NULL 300.00 NULL ...至于列日期标题。我不确定最好的选择是什么。我可以添加一组额外的列，并创建一个可以在结果报告中使用的计算值。
SQL Fiddle：http://www.sqlfiddle.com/#!6/272e5/1/0
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/11215878
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