小编典典

如何将三个SQL选择合并到一个查询中

sql

我下面有三个查询,我想将它们组合成一个查询,这样我就得到了三列按县划分的结果。我尝试对所有表进行内部联接,但数据不正确。如何合并这三个查询并按县分组?

     select [Total DLL Children] = SUM(cd.NumberOfLanguageSpeakers)
 from ClassroomDLL as cd
 inner join Classrooms as c on cd.Classroom_Id = c.Id
 inner join Sites as s on c.Site_Id = s.Id
 inner join Profiles as p on s.Profile_Id = p.Id
 inner join Counties as co on p.County_Id = co.Id
 group by co.Description

 select [Total Children] = (SUM(demo.NumberOfPreschoolers) + SUM(demo.NumberOfToddlers) + SUM(demo.NumberOfInfants))
 from ClassroomDemographics as demo
 inner join Classrooms as c on demo.Classroom_Id = c.Id
 inner join Sites as s on c.Site_Id = s.Id
 inner join Profiles as p on s.Profile_Id = p.Id
 inner join Counties as co on p.County_Id = co.Id
 group By co.Description

 select co.Description from Counties as co
 group by co.Description

阅读 279

收藏
2021-04-15

共1个答案

小编典典

请尝试这个。基本上,每个子查询都需要返回County.Description,然后可以将它们连接在一起。

 SELECT A.Description, B.[Total DLL Children], C.[Total Children]
 FROM (
     select co.Description from Counties as co
     group by co.Description
     ) A
 LEFT JOIN 
     (
         select co.Description, [Total DLL Children] = SUM(cd.NumberOfLanguageSpeakers)
         from ClassroomDLL as cd
         inner join Classrooms as c on cd.Classroom_Id = c.Id
         inner join Sites as s on c.Site_Id = s.Id
         inner join Profiles as p on s.Profile_Id = p.Id
         inner join Counties as co on p.County_Id = co.Id
         group by co.Description
      ) B
      ON A.DESCRIPTION = B.DESCRIPTION 
LEFT JOIN 
      (
         select co.Description, [Total Children] = (SUM(demo.NumberOfPreschoolers) + SUM(demo.NumberOfToddlers) + SUM(demo.NumberOfInfants))
         from ClassroomDemographics as demo
         inner join Classrooms as c on demo.Classroom_Id = c.Id
         inner join Sites as s on c.Site_Id = s.Id
         inner join Profiles as p on s.Profile_Id = p.Id
         inner join Counties as co on p.County_Id = co.Id
         group By co.Description
      ) C
      ON A.DESCRIPTION = C.DESCRIPTION
2021-04-15