select a.cir_flag ,(select count(*) from tab_sub where cir_flag=a.sequence_no) from tab_main a
select a.colname,count(1) from a,b where a.colname=b.colname group by a.colname;
0<>null, null<>0, null<>null
select a.* ,(select count(*) from an_bureau c where 1 = 2) from an_bs a,an_bureau b where a.bureau_code = b.bureau_code
回:bzszp(www.bzszp.533.net)这样只能找出主表和子表都有的记录。 回:zlqs(啊啊啊)你的语句在SQL Server下是正确的,但在Oracle下不执行。 我在SQL Server下的SQL如下: select a.ckfl,isnull((select count(*) from wz_zbfl where ckfl=a.ckfl),0) as zbs from wz_ckfl a
再加一条:要求能在PL/SQL下测试通过
使用外关联: select a.ckfl,nvl(b.ckfl,0) from wz_ckfl a,wz_zbfl b where a.ckfl=b.ckfl(+)
使用外关联: select a.ckfl,nvl(b.ckfl,0) from wz_ckfl a, wz_zbfl b where a.ckfl=b.ckfl(+)
这句没问题: select a.ckfl,(select count(1) from wz_zbfl where ckfl=a.ckfl) zbs from wz_ckfl a
select a.* ,(select nvl(count(*),0) from an_bureau b where a.bureau_code = b.bureau_code) from an_bs a
select count(*) from a,b where a.id=b.id(+)a 主表 b 从表
谢谢shawnzhao(),之所以只给99分,因为完全符合我要求的语句如下: select distinct a.ckfl,a.ckname,nvl(count(b.ckfl),0) from wz_ckfl a, wz_zbfl b where a.ckfl=b.ckfl(+) group by a.ckfl,a.ckname 自己试出来的:)
谢谢shawnzhao(),之所以只给99分,因为完全符合我要求的语句如下: select distinct a.ckfl,a.ckname,nvl(count(b.ckfl),0) from wz_ckfl a, wz_zbfl b where a.ckfl=b.ckfl(+) group by a.ckfl,a.ckname 自己试出来的:)
where a.colname=b.colname group by a.colname;
null<>0,
null<>null
from an_bs a,an_bureau b where a.bureau_code = b.bureau_code
回:zlqs(啊啊啊)你的语句在SQL Server下是正确的,但在Oracle下不执行。
我在SQL Server下的SQL如下:
select a.ckfl,isnull((select count(*) from wz_zbfl where ckfl=a.ckfl),0) as zbs from wz_ckfl a
select a.ckfl,nvl(b.ckfl,0)
from wz_ckfl a,wz_zbfl b
where a.ckfl=b.ckfl(+)
select a.ckfl,nvl(b.ckfl,0)
from wz_ckfl a, wz_zbfl b
where a.ckfl=b.ckfl(+)
select a.ckfl,(select count(1) from wz_zbfl where ckfl=a.ckfl) zbs from wz_ckfl a
from an_bs a
select distinct a.ckfl,a.ckname,nvl(count(b.ckfl),0)
from wz_ckfl a, wz_zbfl b
where a.ckfl=b.ckfl(+) group by a.ckfl,a.ckname
自己试出来的:)
select distinct a.ckfl,a.ckname,nvl(count(b.ckfl),0)
from wz_ckfl a, wz_zbfl b
where a.ckfl=b.ckfl(+) group by a.ckfl,a.ckname
自己试出来的:)