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下面的这个如何写成嵌套的SQL语句
非嵌套的SQL语句：
insert into %s
select A.JcNo, B.QdName, C.CcKindName, Sum(A.Distance) as Distance,
Sum(A.Degree) as Degree, sum(A.Wdistance) as WDistance,
Sum(A.Power) as Power
from %s as A, %s as B, %s as C
where A.QdNo=B.QdNo and A.CcKind=C.CcKindNo
group by A.JcNo,B.QdName,C.CcKindName
order by A.JcNo ASC, B.QdName ASC, C.CcKindName ASC
insert into %s
select A.JcNo, B.QdName, C.CcKindName, Sum(A.Distance) as Distance,
Sum(A.Degree) as Degree, sum(A.Wdistance) as WDistance,
Sum(A.Power) as Power
from %s as A, %s as B, %s as C
where A.QdNo=B.QdNo and A.CcKind=C.CcKindNo
group by A.JcNo,B.QdName,C.CcKindName
order by A.JcNo ASC, B.QdName ASC, C.CcKindName ASC
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licher 2003-12-12
不过，邹建大哥的那种写法有问题？
licher 2003-12-12
楼上老兄：
不知道邹建大哥的写法效率如何？
我是认为嵌套的SQL效率最高，你意见如何？
不知道邹建大哥的写法效率如何？
我是认为嵌套的SQL效率最高，你意见如何？
zjcxc--个人微信公共号同名 2003-12-10
其实楼主的那种写法是效率最高的.
zjcxc--个人微信公共号同名 2003-12-10
select jcno
,qdname=(select qdname from %s where QdNo=a.QdNo)
,CcKindName=(select qdname from %s where CcKind=a.CcKind)
,Sum(A.Distance) as Distance
,Sum(A.Degree) as Degree
,sum(A.Wdistance) as WDistance
,Sum(A.[Power]) as [Power ]
from %s a
group by jcno,qdname,CcKindName
order by JcNo,QdName,CcKindName
,qdname=(select qdname from %s where QdNo=a.QdNo)
,CcKindName=(select qdname from %s where CcKind=a.CcKind)
,Sum(A.Distance) as Distance
,Sum(A.Degree) as Degree
,sum(A.Wdistance) as WDistance
,Sum(A.[Power]) as [Power ]
from %s a
group by jcno,qdname,CcKindName
order by JcNo,QdName,CcKindName
CrazyFor 2003-12-10
什么是嵌套的SQL语句啊?
txlicenhe 2003-12-10
就象楼主的写法挺好啊。
改为JOIN：
insert into %s
select A.JcNo, B.QdName, C.CcKindName, Sum(A.Distance) as Distance,
Sum(A.Degree) as Degree, sum(A.Wdistance) as WDistance,
Sum(A.Power) as Power
from %s as A
join %s as B on A.QdNo=B.QdNo
join %s as C on A.CcKind=C.CcKindNo
--where A.QdNo=B.QdNo and A.CcKind=C.CcKindNo
group by A.JcNo,B.QdName,C.CcKindName
order by A.JcNo ASC, B.QdName ASC, C.CcKindName ASC
改为JOIN：
insert into %s
select A.JcNo, B.QdName, C.CcKindName, Sum(A.Distance) as Distance,
Sum(A.Degree) as Degree, sum(A.Wdistance) as WDistance,
Sum(A.Power) as Power
from %s as A
join %s as B on A.QdNo=B.QdNo
join %s as C on A.CcKind=C.CcKindNo
--where A.QdNo=B.QdNo and A.CcKind=C.CcKindNo
group by A.JcNo,B.QdName,C.CcKindName
order by A.JcNo ASC, B.QdName ASC, C.CcKindName ASC
licher 2003-12-10
中间的：%s分别代表一个表名。