Question

我有两张桌子A和B

date      Fruit_id amount
date_1           1      5
date_1           2      1    
date_2           1      2 
....

date      Fruit_id  amount
date_1           1       2
date_1           3       2    
date_2           5       1 
....

和id_table C. fruit_id fruit 1个苹果 ....

我尝试得到一张桌子，显示某一天每个水果彼此相邻的两张桌子的数量。我试过了

SELECT a.date, f.fruit, a.amount as amount_A, b.amount as amount_B
from table_A a
JOIN table_C f USING(fruit_id)
LEFT JOIN table_B b ON a.date = b.date AND a.fruit_id =  d.fruit_id
WHERE a.date ='myDate'

现在每个水果创建几行而不是1，并且值似乎是相当随机的金额组合。我怎样才能得到一张整洁的桌子

date    fruit  A  B
myDate  apple  1  5
myDate  cherry 2  2
....

Answer 1

尝试

select A.date, C.fruitname, A.amount, B.amount
from A,B,C
where
A.date = B.date
and A.fruitid = B.fruitid
and A.fruitid = C.fruitid
order by A.fruitid

Answer 2

select A.DATE as DATE,
       A.FRUIT_ID FRUIT,
       sum(A.AMOUNT) A,
       sum(B.AMOUNT) B
  from A,B
 where A.DATE = B.DATE
   and A.FRUIT_ID = B.FRUIT_ID
 group by A.DATE,A.FRUIT_ID
 order by A.DATE asc

Answer 3

有几个选项，一个是完全连接，另一个是子查询...出于某种原因，我将首先执行子查询选项。

select x.date, x.fruit_id,
(select a.amount from a where a.fruit_id=x.fruit_id and a.date=x.date) a, 
(select b.amount from b where b.fruit_id=x.fruit_id and b.date=x.date) b
from (select a.date,a.fruit_id from a union select b.date,
b.fruit_id from b) as x

另外你可以使用完整的连接，虽然我没有oracle来测试这个版本，但我相信它会起作用。

select nvl(a.date,b.date) as date, nvl(a.fruit_id,b.fruit_id) as fruit_id,
a.amount as a, b.amount as b
from a full outer join b on b.date=a.date and b.fruit_id=a.fruit_id

Answer 4

我认为您的联接很可能无法返回足够的记录。这个应该返回所有水果，其中A或B中的记录具有匹配量

SELECT ab.date, f.fruit, ab.amount_A, ab.amount_B
table_C f
JOIN ( SELECT a.date, 
              COALESCE(a.fruit_id, b.fruit_id) as fruit_id, 
              COALESCE(a.amount, 0) as amount_A, 
              COALESCE(b.amount, 0) as amount_B
       from table_A a
       FULL OUTER JOIN table_B b ON a.date = b.date AND a.fruit_id =  d.fruit_id
     ) ab ON f.fruit_id = ab.fruit_id

Answer 5

使用您的数据对此进行测试，它似乎有效。您想要使用完整的外部联接以包括A和B中的所有水果和日期：）

SELECT C.Name
     , COALESCE(A.Date, B.Date)
     , COALESCE(A.Amount, 0) 
     , COALESCE(B.Amount, 0)
  FROM C
  JOIN (A FULL OUTER JOIN B ON B.FruitID = A.FruitID AND B.Date = A.Date)
    ON COALESCE(A.FruitID, B.FruitID) = C.FruitID 
 ORDER BY 1, 2

加入几个条件

5 个答案: