使用full join 实现 select * from (查询1的语句) A FULL JOIN (查询2的语句) B ON A.SHOP_TIME=B.SHOP_TIME 想把空替换为0,使用nvl函数即可
当然确定、就是这个样子的、 3楼怎么用NVL啊、full join那不是我想要的效果
倒数第二张图,左右貌似不对第一张表是交易表,第二张是退款,应该是要这个效果吧 select nvl(a.shoptime,b.shoptime),nvl(a.cnt,0), nvl(a.money,0),nvl(b.cnt,0),nvl(b.money,0) from (select tr.shoptime,count(tr.id) cnt,sum(tr.shopmoney) money from tb12_shop_trade tr group by tr.shoptime) a full join (select re.shoptime,count(re.id) cnt,sum(re.shopmoney) money from tb12_shop_reback re group by re.shoptime) b on a.shoptime=b.shoptime
select *
from (查询1的语句) A FULL JOIN (查询2的语句) B
ON A.SHOP_TIME=B.SHOP_TIME
想把空替换为0,使用nvl函数即可
3楼怎么用NVL啊、full join那不是我想要的效果
select nvl(a.shoptime,b.shoptime),nvl(a.cnt,0),
nvl(a.money,0),nvl(b.cnt,0),nvl(b.money,0)
from
(select tr.shoptime,count(tr.id) cnt,sum(tr.shopmoney) money
from tb12_shop_trade tr group by tr.shoptime) a
full join
(select re.shoptime,count(re.id) cnt,sum(re.shopmoney) money
from tb12_shop_reback re group by re.shoptime) b
on a.shoptime=b.shoptime