【发布时间】:2018-05-27 18:45:27
【问题描述】:
【问题讨论】:
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请发布您到目前为止尝试过的内容,发布代码和您遇到的问题。
【问题讨论】:
使用 UNION ALL,您可以从 2 个选择中获得 1 个组合结果集。
然后你可以对它进行分组并对每个日期的金额求和。
所以你可能正在寻找这样的东西:
select
q.ID,
q.Name,
nullif(sum(case when q.Date = '2018-05-01' then q.Amount end), 0) as "5/1/2018",
nullif(sum(case when q.Date = '2018-05-02' then q.Amount end), 0) as "5/2/2018"
from
(
select u1.ID, u1.Name, a1.Date, a1.Amount
from DB1.Table1 AS u1
join DB1.Table2 AS a1 on (a1.ID = u1.ID and a1.Amount is not null)
where a1.Date IN ('2018-05-01', '2018-05-02')
union all -- combines the results of the 2 selects into one resultset
select u2.ID, u2.Name, a2.Date, a2.Amount
from DB2.Table1 AS u2
join DB2.Table2 AS a2 on (a2.ID = u2.ID and a2.Amount is not null)
where a2.Date IN ('2018-05-01', '2018-05-02')
) AS q
group by q.ID, q.Name
order by q.ID;
另一种方法是将它们全部加入。
select
coalesce(a1.ID, a2.ID) as ID,
max(coalesce(u1.Name, u2.Name)) as Name,
max(case
when coalesce(a1.Date, a2.Date) = '2018-05-01'
and coalesce(a1.Amount, a2.Amount) is not null
then coalesce(a1.Amount, 0) + coalesce(a2.Amount, 0)
end) as "5/1/2018",
max(case
when coalesce(a1.Date, a2.Date) = '2018-05-02'
and coalesce(a1.Amount, a2.Amount) is not null
then coalesce(a1.Amount, 0) + coalesce(a2.Amount, 0)
end) as "5/2/2018"
from DB1.Table2 AS a1
full join DB2.Table2 AS a2 on (a2.ID = a1.ID and a2.Date = a1.Date)
left join DB1.Table1 AS u1 on (u1.ID = a1.ID)
left join DB2.Table1 AS u2 on (u2.ID = a2.ID)
where coalesce(a1.Date, a2.Date) IN ('2018-05-01', '2018-05-02')
group by coalesce(a1.ID, a2.ID)
order by coalesce(a1.ID, a2.ID);
但请注意,这样一来,假设两个 Table2 在 (ID, Date) 上具有唯一性
T-Sql 测试数据: 声明@DB1_Table1 表(id int,Name varchar(30)); 声明@DB2_Table1 表(id int,Name varchar(30)); 声明@DB1_Table2 table (id int, [Date] date, Amount decimal(8,2)); 声明@DB2_Table2 table (id int, [Date] date, Amount decimal(8,2)); 插入@DB1_Table1 (id, Name) 值 (1,'Susan'),(2,'Juan'),(3,'Tracy'),(4,'Jenny'),(5,'Bill'); 插入@DB2_Table1 (id, Name) 值 (1,'Susan'),(2,'Juan'),(3,'Tracy'),(4,'Jenny'),(5,'Bill'); 插入@DB1_Table2 (id, [Date], Amount) 值 (1,'2018-05-01',20),(2,'2018-05-01',null),(3,'2018-05-01',30),(4,'2018-05- 01',50),(5,'2018-05-01',null), (1,'2018-05-02',15),(2,'2018-05-02',40),(3,'2018-05-02',25),(4,'2018-05- 02',8),(5,'2018-05-02',null); 插入@DB2_Table2 (id, [Date], Amount) 值 (1,'2018-05-01',null),(2,'2018-05-01',15),(3,'2018-05-01',20),(4,'2018-05- 01',10),(5,'2018-05-01',null), (1,'2018-05-02',15),(2,'2018-05-02',30),(3,'2018-05-02',35),(4,'2018-05- 02',null),(5,'2018-05-02',30);
【讨论】: