LEFT JOIN 3rd Table 到原始表而不是结果表

Question

我有两张表，例如

1. 收入（点、日期、公司）
   
2. 结果（点、日期、出）
   

其中 point 和 date 是主键。
我需要找出任何给定日期的总收入inc 和结果out。 我的回答

WITH all_date AS
(SELECT point, date FROM income
UNION
SELECT point, date FROM outcome)
SELECT a.point, a.date, SUM(inc), SUM(out)
FROM all_date a LEFT JOIN income i ON i.point=a.point AND i.date=d.date
                LEFT JOIN outcome o ON o.point=a.point AND o.date=a.date
GROUP BY a.point, a.date

但是我得到了错误的结果。因为all_date LEFT JOIN income 在尝试使用outcome 执行LEFT JOIN 时会复制income 中的某些行，因此会弄乱最终的SUM。

基本上我想LEFT JOIN 第三张桌子outcome 和all_date 不是 all_date LEFT JOIN income。请参阅http://www.sql-ex.ru/learn_exercises.php 的 Qus No 30. 以获得进一步的说明

Answer 1

那为什么不先加入呢？

或者更好的是，只做一个完整的外连接：

Select coalesce(o.point, i.point) point,
   coalesce(o.Date, i.date) date
From income i full join outcome o
  on o.point=i.point 
     and o.date=i.date
group by coalesce(o.point, i.point),
     coalesce(o.Date, i.date)

而且，由于您实际上没有使用任何聚合函数，因此 distinct 也可以：

Select distinct
   coalesce(o.point, i.point) point,
   coalesce(o.Date, i.date) date
From income i full join outcome o
  on o.point=i.point 
     and o.date=i.date

抱歉，我不是 MySQL 专家，也没有意识到 MySQL 没有 Full Join 语法。但有一个解决方法。请看以下链接：

Full Join in MYSQL

基本上你需要一个联合，（实际上接近你所做的）

select point, date 
from income i
 left join outcome o 
    on o.point=i.point 
      and o.date=i.date  
  UNION                 -- <-- leave out the ALL to eliminate duplicates
select point, date 
from outcome o
 right join income i 
    on i.point=o.point 
      and i.date=o.date  

Answer 2

这是我在 MSSQL 中的最终解决方案。但我想使用COALESCE 和FULL JOIN 效果最好。

    WITH all_date AS
    (SELECT point, date FROM income
    UNION
    SELECT point, date FROM outcome)

    SELECT a.point, a.date,
    (SELECT SUM(o.out) FROM outcome o WHERE o.point=a.point AND o.date=a.date) AS outcome,
    (SELECT SUM(i.inc) FROM income i WHERE i.point=a.point AND i.date=a.date) AS income

    FROM all_date a

Answer 3

MS-SQL sp 解决您的问题：

CREATE PROCEDURE dototals
@date DATETIME
AS
BEGIN

SELECT point,SUM(ISNULL(inc,0)) as inc into #tmp_i FROM Income WHERE date =@date  GROUP BY point

SELECT point,SUM(ISNULL(out,0)) as out into #tmp_d FROM Outcome WHERE date =@date  GROUP BY point

SELECT DISTINCT point into #tmp_p FROM 
(SELECT point FROM #tmp_i
UNION ALL
SELECT point FROM #tmp_o)

SELECT @date as date, t1.point as point, t2.out as outcome,t3.inc as income 
FROM #tmp_p as t1
LEFT JOIN #tmp_o as t2 on t1.point=t2.point
LEFT JOIN #tmp_i as t3 on t1.point=t3.point
END

Answer 4

不知道这里用表表达式的目的是什么，这里用一个join就可以得到结果：

SELECT i.point, i.date, SUM(inc), SUM(out)
FROM  income i 
LEFT JOIN outcome o   
ON o.point=i.point AND o.date=i.date
GROUP BY i.point, i.date