我有一个订单表和一个包含付款信息的行程表。这是一个多对多的关系——订单可以分成许多次旅行,一次旅行可以有几个订单的付款信息,有时是一个订单。Trips表中没有"零"记录,因此以该值作为键的左连接将返回NULL记录。我正在使用SQL 2012
Order table
+----+----------+--------------+
| order_id | trip_nbr | veh_id |
+----+----------+--------------+
| 1 | 12 | 3 |
| 2 | 22 | 6 |
| 2 | 0 | 8 |
| 4 | 25 | 7 |
| 7 | 0 | 11 |
+----+----------+--------------+
Trips table
+----+------------+--------------+
| trip_nbr | payment | veh_id |
+----+------------+--------------+
| 12 | 20.00 | 3 |
| 22 | 123.00 | 6 |
| 22 | 12.50 | 6 |
| 25 | 133.33 | 7 |
+----+------------+--------------+
这是我的查询:
select o.order_id,
t.trip_nbr,
sum(t.payment_amt)
from orders o
left outer join trips t
on o.trip_nbr = t.trip_nbr
group by o.order_id,
t.trip_nbr
结果:
+----+----------+--------------+
| order_id | trip_nbr | sum |
+----+----------+--------------+
| 1 | 12 | 20.00 |
| 2 | 22 | 135.50 |
| 2 | NULL | NULL |
| 4 | 25 | 133.33 |
| 7 | NULL | NULL |
+----+----------+--------------+
问题是,我从订单表中获得了很多信息,而从Trips表中只获得了付款信息。所以我不想排除任何订单记录(这将发生,如果我添加子句"WHERE t.t trip_nbr不是NULL")-但我不想在我的分组中得到2条记录-一个为t.t trip_nbr是NULL,一个在那里它找到一个匹配。
所需结果:+----+----------+--------------+
| order_id | trip_nbr | sum |
+----+----------+--------------+
| 1 | 12 | 20.00 |
| 2 | 22 | 135.50 |
| 4 | 25 | 133.33 |
| 7 | NULL | NULL |
+----+----------+--------------+
我希望不匹配的记录order_id = 2被"汇总"-但保留order_id = 7的单独记录。原因是这个表稍后将与另一个表连接,并且额外的NULL记录正在创建重复项。
应该可以:
WITH orders2 AS
(
SELECT *,
N = SUM(CASE WHEN trip_nbr <> 0 THEN 1 ELSE 0 END) OVER(PARTITION BY order_id)
FROM orders
)
SELECT o.order_id,
t.trip_nbr,
SUM(t.payment_amt)
FROM orders2 o
LEFT OUTER JOIN trips t
ON o.trip_nbr = t.trip_nbr
WHERE N = 0 OR (N > 1 AND o.trp_nbr <> 0)
GROUP BY o.order_id,
t.trip_nbr;
您可以使用像RANK
这样的窗口函数来识别多余的NULL
值记录,并在外部查询中过滤它们:
select order_id,
trip_nbr,
total_payment
from (
select o.order_id,
t.trip_nbr,
sum(t.payment) as total_payment,
rank() over (partition by order_id
order by case
when t.trip_nbr IS NULL then 2
else 1
end) as rnk
from orders o
left outer join trips t
on o.trip_nbr = t.trip_nbr
group by o.order_id, t.trip_nbr) as t
where t.rnk = 1
如果将空值转换为零,则对给定order_id的"trip_nbr"one_answers"sum"求和。这难道不能解决你的挑战吗?
create table #Order (Order_Id int , Trip_nbr int , Veh_id int )
Create Table #Trips (trip_nbr int , Payment Numeric(13,2), Veh_id int )
insert into #Order (Order_id, Trip_nbr, Veh_id) values (1,12,3)
insert into #Order (Order_id, Trip_nbr, Veh_id) values (2,22,6)
insert into #Order (Order_id, Trip_nbr, Veh_id) values (2,0 ,8)
insert into #Order (Order_id, Trip_nbr, Veh_id) values (4,25,7)
insert into #Order (Order_id, Trip_nbr, Veh_id) values (7,0,11)
insert into #Trips (trip_nbr, Payment, Veh_id) values (12, 20.00 , 3 )
insert into #Trips (trip_nbr, Payment, Veh_id) values (22, 123.00,6 )
insert into #Trips (trip_nbr, Payment, Veh_id) values (22, 12.50 , 6 )
insert into #Trips (trip_nbr, Payment, Veh_id) values (25, 133.33 , 7 )
select Order_id, trip_nbr = sum(trip_nbr), Payment = sum(payment)
from
(
select o.order_id,
t.trip_nbr,
Payment = sum(t.Payment)
from #order o
left outer join #trips t on t.trip_nbr = o.trip_nbr
-- left outer join #order o on t.trip_nbr = o.trip_nbr
group by o.order_id, t.trip_nbr
) x
group by Order_id
order by Order_id