Question

SELECT 
    Income.point, Income.date, SUM(out), SUM(inc)
FROM 
    Income 
LEFT JOIN
    Outcome ON Income.point = Outcome.point 
           AND Income.date = Outcome.date
GROUP BY 
    Income.point, Income.date

UNION

SELECT 
    Outcome.point, Outcome.date, SUM(out), SUM(inc)
FROM 
    Outcome 
LEFT JOIN
    Income ON Income.point = Outcome.point 
           AND Income.date = Outcome.date
GROUP BY 
    Outcome.point, Outcome.date;

我有这个代码，我想要做的是在加入之前分组。 “假设我们有一个包含连接和分组的SQL查询。评估此类查询的标准方法是首先执行所有连接，然后执行分组操作。但是，可以执行组 - 早期，即推动分组操作超过一个或多个连接。早期分组可以通过减少参与连接的数据量来降低查询处理成本。“ 所以我需要解释如何做到这一点

在这种情况下，

运动如下：

假设每个出口（点）的钱的收入（inc）和费用（out）每天登记任意次数，得到的结果集包括：出口，日期，费用，收入。

请注意，单个记录必须与每个日期的每个商店对应。

使用收入和结果表。

Answer 1

试试此代码

SELECT ip,id,ii,oo FROM
(SELECT I.point ip, I.date id, SUM(I.inc) ii FROM Income I GROUP BY I.point, I.date ) in1
LEFT JOIN
(SELECT O.point op, O.date od, SUM(O.out) oo FROM Outcome O GROUP BY O.point, O.date ) ou1 
ON op=ip AND od=id

UNION

SELECT ip,id,ii,oo FROM
(SELECT I.point ip, I.date id, SUM(I.inc) ii FROM Income I GROUP BY I.point, I.date ) in1
RIGHT JOIN
(SELECT O.point op, O.date od, SUM(O.out) oo FROM Outcome O GROUP BY O.point, O.date ) ou1 
ON op=ip AND od=id

也许有人可以给它起个名字。我甚至不知道你如何在括号中称这些SELECTS ......： - /

修改

那么，考虑到Luis LL的想法并将其与“早期分组”相结合，可以获得以下内容：

SELECT COALESCE(ip,op) point,COALESCE(id,od) date,ii inc,oo out FROM (SELECT point ip, date id, SUM(inc) ii FROM Income GROUP BY point, date ) in1 FULL OUTER JOIN (SELECT point op, date od, SUM(out) oo FROM Outcome GROUP BY point, date ) ou1 ON op=ip AND od=id

也许这会成功吗？

Answer 2

我不知道你想要达到什么目标，但在我看来，你需要的是FULL OUTER JOIN

SELECT 
    ISNULL(Income.point, OutCome.point) AS Point, ISNULL(Income.date, OutComeDate) AS [date], SUM(out) AS Expenses, SUM(inc) AS Income
FROM Income 
FULL OUTER JOIN Outcome 
    ON Income.point = Outcome.point AND Income.date = Outcome.date
GROUP BY Income.point, Income.date

Answer 3

如果我正确理解您的问题，您可以使用CTE。

WITH CTE
AS
(
    SELECT [point]
    ,      [date]
    ,      [SumOut] = SUM([out])
    ,      [SumInc] = SUM([inc])
    FROM Income 
    GROUP BY [point], [date]
)

SELECT CTE.[Point]
,      CTE.[Date]
,      CTE.[SumOut]
,      CTE.[SumInc]
FROM CTE
LEFT JOIN Outcome ON CTE.[point] = Outcome.point AND CTE.[date] = Outcome.[date]

Answer 4

假设您需要在2个单独的表中注册此数据（不是最优雅的方式），您可以使用子查询。 UNION可能不是这样做的，因为它不会将每个“点”的数据集放在一个记录中。

Answer 5

SELECT A. * FROM （SELECT Income.point，Income.date，SUM（out），SUM（inc）从收入 LEFT JOIN 收入来自Income.point = Outcome.point AND Income.date = Outcome.date 通过...分组 Income.point，Income.date）A

UNION

SELECT B. * FROM （SELECT Outcome.point，Outcome.date，SUM（out），SUM（inc）从结果 LEFT JOIN 收入ON Income.point = Outcome.point AND Income.date = Outcome.date 通过...分组 Outcome.point，Outcome.date）B; </ p>

Answer 6

Select  ip,id,oo,ii from(
Select i.point as ip,i.date as id,sum(i.inc) as ii from Income as I group BY i.point, i.date) in1

Left join (
Select o.point as op, o.date as od,sum(o.out) as oo from outcome as O group by o.point, o.date) ou1
on ip=op and id=od

union

Select  op,od,oo,ii from(
Select o.point as op, o.date as od,sum(o.out) as oo from outcome as O group by o.point, o.date) ou2

Left join (
Select i.point as ip,i.date as id,sum(i.inc) as ii from Income as I group BY i.point, i.date) in2
on ip=op and id=od

JOIN之前的SQL GROUP BY - 查询时的序列

6 个答案: