通过以下SQL语句查询:
select mealname,COUNT(*) 份数 from Melinfo2 group by mealname
select mealname,sum(份数) 份数 from tb group by mealname
得到下面两个查询结果
表1, mealname 份数 表2. mealname 份数
鸡蛋饭 5 肉段饭 3
排骨饭 2 鸡蛋饭 3
如何能然上边两个表变成一个表啊,就像表3一样
表3,mealname 份数
鸡蛋饭 8
排骨饭 2
肉段饭 3
跪求高手解决~~~
select mealname,COUNT(*) 份数 from Melinfo2 group by mealname
select mealname,sum(份数) 份数 from tb group by mealname
得到下面两个查询结果
表1, mealname 份数 表2. mealname 份数
鸡蛋饭 5 肉段饭 3
排骨饭 2 鸡蛋饭 3
如何能然上边两个表变成一个表啊,就像表3一样
表3,mealname 份数
鸡蛋饭 8
排骨饭 2
肉段饭 3
跪求高手解决~~~
select t.mealname,sum(t.份数) 份数
from (
select mealname,COUNT(*) 份数 from Melinfo2 group by mealname
UNION ALL
select mealname,sum(份数) 份数 from tb group by mealname
)
t group by t.mealname
select mealname,sum(份数) from (
select mealname,COUNT(*) 份数 from Melinfo2 group by mealname
union all
select mealname,sum(份数) 份数 from tb group by mealname) as t group by mealname
select mealname,sum(份数) from (
select mealname,COUNT(*) 份数 from Melinfo2 where datediff(day,datetime,getdate())=0 group by mealname
union all
select mealname,sum(份数) 份数 from tb where datediff(day,datetime,getdate())=0 group by mealname) as t group by mealname
from (
select mealname,COUNT(*) 份数 from Melinfo2 group by mealname
UNION ALL
select mealname,sum(份数) 份数 from tb group by mealname
)
t
where
convert(datetime,convert(char(20),[maketime],102))= '" + t +"'
group by
t.mealname
鉴于你已去到了结果集“select mealname,COUNT(*) 份数 from Melinfo2 group by mealname
union all
select mealname,sum(份数) 份数 from tb group by mealname) as t group by mealname
”
不用聚合函数也是可以的;可以写成如下:
“SELECT T.mealName,sum(T.mealNum) AS 份数 FROM
(
SELECT mealName,mealNum FROM tb1
UNION ALL
SELECT mealName,mealNum FROM tb2
)T
--where T.mealName=N'鸡蛋饭'(如需带where条件,在group by前加上条件)
group by T.mealName”