Question

如果summ为正+，则表示帐户拥有资金，如果记录为负-，则表示帐户有付款。我需要找出今天哪个帐户没有任何资金，我有这个问题：

SELECT  a.Customer
       ,a.Deal
       ,(a.positive + b.negative) AS own_to_the_bank
FROM    (
          SELECT    SUM(Summ) AS positive
                   ,Customer
                   ,Deal
          FROM      #test
          WHERE     Summ > 0
          GROUP BY  Customer
                   ,Deal
        ) AS a 
JOIN    (
          SELECT    SUM(Summ) AS negative
                   ,Customer
                   ,Deal
          FROM      #test
          WHERE     Summ < 0
          GROUP BY  Customer
                   ,Deal
        ) AS b ON a.Customer = b.Customer
                  AND a.Deal = b.Deal
WHERE a.positive + b.negative >0

及其工作正常，所以现在我必须找出当我的查询a.positive + b.negative = 0时帐户停止向银行取款的时候。坚持这个问题几个小时，任何帮助？

Answer 1

我开始创建每日余额，客户，交易和货币

SELECT t1.Customer, t1.Deal, t1.Currency, t1.Date, Balance = (SELECT SUM(Summ) FROM #test as hist WHERE hist.Customer = t1.Customer and hist.Deal = t1.Deal and hist.Currency = t1.Currency and hist.Date <= t1.Date)
FROM #test as t1

添加了正余额和rownum条件（按日期排序）

SELECT Customer, Deal, Currency, Date, Balance, RowNum = ROW_NUMBER() OVER(PARTITION BY Customer, Deal, Currency ORDER BY Date) 
FROM
(
            select t1.Customer, t1.Deal, t1.Currency, t1.Date, Balance = (SELECT SUM(Summ) FROM #test as hist WHERE hist.Customer = t1.Customer and hist.Deal = t1.Deal and hist.Currency = t1.Currency and hist.Date <= t1.Date)
            FROM #test as t1
             ) as inn
WHERE Balance > 0

最后选了第一个。

SELECT Customer, Deal, Currency, Date, Balance
FROM (  SELECT Customer, Deal, Currency, Date, Balance, RowNum = ROW_NUMBER() OVER(PARTITION BY Customer, Deal, Currency ORDER BY Date) 
        FROM
        (
                    SELECT t1.Customer, t1.Deal, t1.Currency, t1.Date, Balance = (SELECT SUM(Summ) FROM #test as hist WHERE hist.Customer = t1.Customer and hist.Deal = t1.Deal and hist.Currency = t1.Currency and hist.Date <= t1.Date)
                    FROM #test as t1
                     ) as t
        WHERE Balance > 0 ) as t2
WHERE t2.RowNum = 1

Answer 2

当客户停止时，您可能会有几个日期

例如，我们有两个日期：

此查询显示最后一个：

select distinct a.customer, a.date
from test as a 
left join test as b 
on a.Date > b.Date and a.Customer = b.Customer 
where a.summ < 0 and b.summ > 0 
group by a.customer order by a.date, b.date desc

线索是按日期在不同方向订购连接表，然后按每个客户的第一行排序。

帐户成为时需要找出日期

2 个答案: