如果状态为'Y',我想加入这些表并计数
table1(date, status)
8/23/2015 Y
8/24/2015 Y
8/24/2015 N
table2(date, status)
8/23/2015 Y
8/23/2015 Y
table3(date, status)
8/23/2015 Y
8/25/2015 N
8/25/2015 Y
我期待的结果是。
DATE count(table1.status) count(table2.status) count(table3.status)
--------- -------------------- -------------------- --------------------
8/23/2015 1 2 1
8/24/2015 1 0 0
8/25/2015 0 0 1
也许最简单的方法是将表union all
放在一起,然后进行聚合:
select date, sum(status1) as status1, sum(status2) as status2,
sum(status3) as status3
from ((select date, 1 as status1, 0 as status2 , 0 as status3
from table1
where status = 'Y') union all
(select date, 0 as status1, 1 as status2 , 0 as status3
from table2
where status = 'Y') union all
(select date, 0 as status1, 0 as status2 , 1 as status3
from table3
where status = 'Y')
) t
group by date
order by date;
如果你想用full join
做这件事,你必须非常小心。你很想写:
select date,
sum(case when t1.status1 = 'Y' then 1 else 0 end) as status1,
sum(case when t2.status1 = 'Y' then 1 else 0 end) as status2,
sum(case when t3.status1 = 'Y' then 1 else 0 end) as status3
from table1 t1 full join
table2 t2
using (date) full join
table3 t3
using (date)
group by date
order by date;
但是,当在不同的表中对同一日期有多个计数时(日期的笛卡尔乘积(,就会出现问题。因此,下一个诱惑是添加count(distinct)
。在这种情况下可以这样做,因为没有唯一的列。即使有,这也会增加开销。
最后,如果您想走这条路,可以通过预聚合每个表来解决这个问题。
这里有另一种选择:
with table1 as (select to_date('23/08/2015', 'dd/mm/yyyy') dt, 'Y' status from dual union all
select to_date('24/08/2015', 'dd/mm/yyyy') dt, 'Y' status from dual union all
select to_date('25/08/2015', 'dd/mm/yyyy') dt, 'N' status from dual),
table2 as (select to_date('23/08/2015', 'dd/mm/yyyy') dt, 'Y' status from dual union all
select to_date('23/08/2015', 'dd/mm/yyyy') dt, 'Y' status from dual union all
select to_date('26/08/2015', 'dd/mm/yyyy') dt, 'Y' status from dual),
table3 as (select to_date('23/08/2015', 'dd/mm/yyyy') dt, 'Y' status from dual union all
select to_date('25/08/2015', 'dd/mm/yyyy') dt, 'Y' status from dual union all
select to_date('25/08/2015', 'dd/mm/yyyy') dt, 'N' status from dual)
select coalesce(t1.dt, t2.dt, t3.dt) dt,
coalesce(t1.cnt, 0) t1_cnt,
coalesce(t2.cnt, 0) t2_cnt,
coalesce(t3.cnt, 0) t3_cnt
from (select dt, count(case when status = 'Y' then 1 end) cnt
from table1
group by dt) t1
full outer join (select dt, count(case when status = 'Y' then 1 end) cnt
from table2
group by dt) t2 on t1.dt = t2.dt
full outer join (select dt, count(case when status = 'Y' then 1 end) cnt
from table3
group by dt) t3 on t1.dt = t3.dt
order by coalesce(t1.dt, t2.dt, t3.dt);
DT T1_CNT T2_CNT T3_CNT
---------- ---------- ---------- ----------
23/08/2015 1 2 1
24/08/2015 1 0 0
25/08/2015 0 0 1
26/08/2015 0 1 0
(在连接到其他表之前,它将行聚合到每个dt一行,这样计数中就不会包含重复的行(。
这是一个有趣的问题。
为了得到结果,我们可以使用下面的解码函数对过滤后的"Y"值求和。
create table table1(date1 date, status varchar2(2))
insert into table1 values ( to_date( '8/23/2015', 'mm/dd/yyyy'), 'Y');
insert into table1 values ( to_date( '8/24/2015', 'mm/dd/yyyy'), 'Y');
insert into table1 values ( to_date( '8/24/2015', 'mm/dd/yyyy'), 'N');
create table table2(date1 date, status varchar2(2))
insert into table2 values ( to_date( '8/23/2015', 'mm/dd/yyyy'), 'Y');
insert into table2 values ( to_date( '8/23/2015', 'mm/dd/yyyy'), 'Y');
create table table3(date1 date, status varchar2(2))
insert into table3 values ( to_date( '8/23/2015', 'mm/dd/yyyy'), 'Y');
insert into table3 values ( to_date( '8/25/2015', 'mm/dd/yyyy'), 'N');
insert into table3 values ( to_date( '8/25/2015', 'mm/dd/yyyy'), 'Y');
select date1,
sum(decode(x, '1', status, null)), -- table1:x=1
sum(decode(x, '2', status, null)), -- table2:x=2
sum(decode(x, '3', status, null)) -- table3:x=3
from (select 1 x, date1, decode(status, 'Y', 1, 0) status
from table1
union all
select 2 x, date1, decode(status, 'Y', 1, 0) status
from table2
union all
select 3 x, date1, decode(status, 'Y', 1, 0) status
from table3)
group by date1
order by 1
在@Gordon Linoff解决方案中,他创建了3列(status1、status2、status3(来获得结果,在这个解决方案中我直接从数据中获得结果。
所有的路都通向罗马。
也检查
select t1.DATE,
SUM(NVL(decode(t1.status, 'Y', 1, 0)),0) table1_sum,
SUM(NVL(decode(t2.status, 'Y', 1, 0)),0) table2_sum,
SUM(NVL(decode(t3.status, 'Y', 1, 0)),0) table3_sum
from table1 t1,
join table2 t2,
join table3 t3,
where t1.Date = t2.date and t1.Date = t3.date
group by t1.DATE
order by t1.DATE