Question

我正在尝试对此表格进行查询：

Id       startdate     enddate       amount
1        2013-01-01   2013-01-31      0.00
2        2013-02-01   2013-02-28      0.00
3        2013-03-01   2013-03-31      245
4        2013-04-01   2013-04-30      529
5        2013-05-01   2013-05-31      0.00
6        2013-06-01   2013-06-30      383
7        2013-07-01   2013-07-31      0.00
8        2013-08-01   2013-08-31      0.00

我想得到输出：

2013-01-01          2013-02-28          0
2013-03-01          2013-06-30          1157
2013-07-01          2013-08-31           0

我想得到那个结果，所以我知道什么时候开始进货以及什么时候停止。我也对钱开始进入之前的几个月（第一行解释）以及资金停止的月数感兴趣（这也解释了为什么我对2013年7月到2013年8月的第3行感兴趣）。

我知道我可以在日期上使用最小值和最大值并对金额求和但我无法弄清楚如何将记录划分为这种方式。
谢谢！

Answer 1

with CT as
(
    select t1.*,
           ( select max(endDate) 
             from t 
             where startDate<t1.StartDate and SIGN(amount)<>SIGN(t1.Amount)
           ) as GroupDate
    from t  as t1
) 
select min(StartDate) as StartDate,
       max(EndDate) as EndDate,
       sum(Amount) as Amount
from CT
group by GroupDate
order by StartDate

SQLFiddle demo

Answer 2

这是一个想法（和a fiddle一起使用）：

;WITH MoneyComingIn AS
(
    SELECT MIN(startdate) AS startdate, MAX(enddate) AS enddate, 
        SUM(amount) AS amount
    FROM myTable
    WHERE amount > 0
)
SELECT MIN(startdate) AS startdate, MAX(enddate) AS enddate, 
    SUM(amount) AS amount
FROM myTable
WHERE enddate < (SELECT startdate FROM MoneyComingIn)
UNION ALL
SELECT startdate, enddate, amount
FROM MoneyComingIn
UNION ALL
SELECT MIN(startdate) AS startdate, MAX(enddate) AS enddate, 
    SUM(amount) AS amount
FROM myTable
WHERE startdate > (SELECT enddate FROM MoneyComingIn)

还有一秒，没有使用UNION（fiddle）：

SELECT MIN(startdate), MAX(enddate), SUM(amount)
FROM
(
    SELECT startdate, enddate, amount,
    CASE 
        WHEN EXISTS(SELECT 1 
                    FROM myTable b 
                    WHERE b.id>=a.id AND b.amount > 0) THEN
            CASE WHEN EXISTS(SELECT 1 
                             FROM myTable b 
                             WHERE b.id<=a.id AND b.amount > 0) 
                 THEN 2 
                 ELSE 1 
            END
        ELSE 3
    END AS partition_no
    FROM myTable a
) x
GROUP BY partition_no

虽然我认为正如所写，它假定Id是有序的。您可以使用ROW_NUMBER() OVER(ORDER BY startdate)替换它。

Answer 3

这样的事情应该这样做：

select min(startdate), max(enddate), sum(amount) from paiements
   where enddate   < (select min(startdate) from paiements where amount >0)
union
select min(startdate), max(enddate), sum(amount) from paiements
   where startdate >= (select min(startdate) from paiements where amount >0)
     and enddate   <= (select max(enddate) from paiements where amount >0)
union
select min(startdate), max(enddate), sum(amount) from paiements
   where startdate > (select max(enddate) from paiements where amount >0)

但是对于这种报告，使用多个查询可能更明确。

Answer 4

这样做你想要的：

-- determine the three periods
DECLARE @StartMoneyIn INT
DECLARE @EndMoneyIn INT

SELECT @StartMoneyIn = MIN(Id)
FROM [Amounts]
WHERE amount > 0

SELECT @EndMoneyIn = MAX(Id)
FROM [Amounts]
WHERE amount > 0

-- retrieve the amounts
SELECT MIN(startdate) AS startdate, MAX(enddate) AS enddate, SUM(amount) AS amount
FROM [Amounts]
WHERE Id < @StartMoneyIn
UNION
SELECT MIN(startdate), MAX(enddate), SUM(amount)
FROM [Amounts]
WHERE Id >= @StartMoneyIn AND Id <= @EndMoneyIn
UNION
SELECT MIN(startdate), MAX(enddate), SUM(amount)
FROM [Amounts]
WHERE Id > @EndMoneyIn

Answer 5

如果你想要做的就是看钱什么时候开始进来以及什么时候停止，这可能对你有用：

select 
    min(startdate),
    max(enddate),
    sum(amount)
where
    amount > 0

这不包括没有钱进入的时期。

Answer 6

如果您不关心期间的总数，但只想要从0到某事的记录，反之亦然，那么您可以做一些像这样的疯狂事情：

select *
from MoneyTable mt
where exists ( select *
               from MoneyTable mtTemp
               where mtTemp.enddate = dateadd(day, -1, mt.startDate)
               and mtTemp.amount <> mt.amount
               and mtTemp.amount * mt.amount = 0)

或者如果您必须包含第一条记录：

select *
from MoneyTable mt
where exists ( select *
               from MoneyTable mtTemp
               where mtTemp.enddate = dateadd(day, -1, mt.startDate)
               and mtTemp.amount <> mt.amount
               and mtTemp.amount * mt.amount = 0 )
or not exists ( select *
                from MoneyTable mtTemp
                where mtTemp.enddate = dateadd(day, -1, mt.startDate))

Sql Fiddle

按日期范围分组在t-sql中

6 个答案: