这是一个只会通过dbo.Activity
ONCE的查询
SELECT H.CUSTOMER
,H.LEDGER
,H.ACCOUNT
,MAX(H.ACTIVITY_DATE) ACTIVITY_DATE
,SUM(CASE X.I WHEN 1 THEN AMOUNT ELSE -AMOUNT END) AMOUNT
FROM (SELECT CUSTOMER
,LEDGER
,ACCOUNT
,ACTIVITY_DATE
,AMOUNT
,ROW_NUMBER() OVER (PARTITION BY CUSTOMER, LEDGER, ACCOUNT ORDER BY ACTIVITY_DATE DESC) AS ROW_NUMBER
FROM dbo.ACTIVITY WITH (NOLOCK)
) H
CROSS JOIN (select 1 union all select 2) X(I)
WHERE ROW_NUMBER - X.I >= 0
GROUP BY H.CUSTOMER
,H.LEDGER
,H.ACCOUNT
,ROW_NUMBER - X.I;
这是我用来测试的一些数据的 DDL/DML
CREATE TABLE dbo.ACTIVITY(CUSTOMER int, LEDGER int, ACCOUNT int, ACTIVITY_DATE datetime, AMOUNT int)
INSERT dbo.ACTIVITY select
1,2,3,GETDATE(),123 union all select
1,2,3,GETDATE()-1,16 union all select
1,2,3,GETDATE()-2,12 union all select
1,2,3,GETDATE()-3,1 union all select
4,5,6,GETDATE(),1000 union all select
4,5,6,GETDATE()-6,123 union all select
7,7,7,GETDATE(),99;
备择方案
使用子查询获取上一行的更传统方法:
SELECT CUSTOMER, LEDGER, ACCOUNT, ACTIVITY_DATE,
AMOUNT - ISNULL((SELECT TOP(1) I.AMOUNT
FROM dbo.ACTIVITY I
WHERE I.CUSTOMER = O.CUSTOMER
AND I.LEDGER = O.LEDGER
AND I.ACCOUNT = O.ACCOUNT
AND I.ACTIVITY_DATE < O.ACTIVITY_DATE
ORDER BY I.ACTIVITY_DATE DESC), 0) AMOUNT
FROM dbo.ACTIVITY O
ORDER BY CUSTOMER, LEDGER, ACCOUNT, ACTIVITY_DATE;
或 ROW_NUMBER() 两次数据并在它们之间加入
SELECT A.CUSTOMER, A.LEDGER, A.ACCOUNT, A.ACTIVITY_DATE,
A.AMOUNT - ISNULL(B.AMOUNT,0) AMOUNT
FROM (SELECT *, RN=ROW_NUMBER() OVER (partition by CUSTOMER, LEDGER, ACCOUNT
order by ACTIVITY_DATE ASC)
FROM dbo.ACTIVITY) A
LEFT JOIN (SELECT *, RN=ROW_NUMBER() OVER (partition by CUSTOMER, LEDGER, ACCOUNT
order by ACTIVITY_DATE ASC)
FROM dbo.ACTIVITY) B ON A.CUSTOMER = B.CUSTOMER
AND A.LEDGER = B.LEDGER
AND A.ACCOUNT = B.ACCOUNT
AND B.RN = A.RN-1 -- prior record
ORDER BY A.CUSTOMER, A.LEDGER, A.ACCOUNT, A.ACTIVITY_DATE;