紧急求助：如何将下面的结果放到同一张表中?

[libjprayerlibj]

select (len(name)-len(replace(name,''北京'')))/len('北京') name from table7
select (len(home)-len(replace(home,'南京','')))/len('南京') home from table7

结果显示： name home
1 5 1 3
2 3 2 4

请问，如何在存储过程中把上面两张表中的内容（即列 name 与 home 的数据 ） 放到同一张表中？
注释：上面的语句中，查询的参数只能为一个

[miller3000]

没有明白你的意思

[guxiaobo1982]

select (len(name)-len(replace(name,''北京'')))/len('北京') name from table7
union
select (len(home)-len(replace(home,'南京','')))/len('南京') home from
table7 into (你要的表名)

[leimin]

select (len(name)-len(replace(name,'北京','')))/len('北京') as name ,(len(home)-len(replace(home,'南京','')))/len('南京') as home from table7

[zjcxc]

select * into 生成的表
from(
select (len(name)-len(replace(name,''北京'')))/len('北京') name from table7
union all
select (len(home)-len(replace(home,'南京','')))/len('南京') home from table7
) a

[zjcxc]

如果表已经存在:
insert into 表
select * from(
select (len(name)-len(replace(name,''北京'')))/len('北京') name from table7
union all
select (len(home)-len(replace(home,'南京','')))/len('南京') home from table7
) a

[zjcxc]

如果你要生成横向连接的表,就用临时做中间处理过程:

select id=identity(int,1,1),(len(name)-len(replace(name,''北京'')))/len('北京') name
into #tb1 from table7

select id=identity(int,1,1),(len(home)-len(replace(home,'南京','')))/len('南京') home
into #tb2 from table7

--合并结果
select a.name,b.home into 要生成的表
from #tb1 a full join #tb2 b on a.id=b.id

或:
insert into 表
select a.name,b.home from #tb1 a full join #tb2 b on a.id=b.id

--删除临时表
drop table #tb1,#tb2

[prayeroihll]

select (len(name)-len(replace(name,''北京'')))/len('北京') name from table7
union
select (len(home)-len(replace(home,'南京','')))/len('南京') home from
table7 into (你要的表名)
应该就可以啦，你试一试啊！

[yczzg]

select a.ID,a.name,b.home
from
(select ID,(len(name)-len(replace(name,''北京'')))/len('北京') name
from table7 )a
inner join (
select ID,(len(home)-len(replace(home,'南京','')))/len('南京') home
from table7)b on a.ID=b.ID

[libjprayerlibj]

求并集？那么，求交集用什么啊？

[fengyun6]

用union ...into...语句

[yujohny]

建一个表aa，设ID自动增加，还有两个字段name,home
然后这样写代码：
INSERT aa (name,home)
SELECT (len(name)-len(replace(name,''北京'')))/len('北京'),
(len(home)-len(replace(home,'南京','')))/len('南京')
FROM table7