Question

我有两个查询，每个查询返回从一堆其他表中的数据派生的相当复杂的表，我想将它们连接在一起，然后在超级连接表上使用group by。

我不认为我可以使用单个查询派生此表，因为每个查询都访问一组不同的表，而排序故事是因为它无法加入所需的所有信息两者都在一排（至少我不能想到一种方式）。

第一个查询是：

select
    pr.runName,
    cp.firstname,
    mp.name
from
    passrun as pr,
    passrunpoly as prp,
    mappolygon as mp,
    cmnemployee as ce,
    cmnperson as cp,
    passschedule as ps

where 
    pr.runid = prp.runid
    and prp.polyid = mp.polyid
    and pr.employeeid = ce.employeeid
    and ce.personid = cp.personid
    and pr.scheduleid = ps.scheduleid
    and ps.ldate = 20170403

第二个查询是：

select 
    mp.name,
    count(distinct pbl.lat) as Stops,
    count(case when pba.spacetype = 'S' then pb.ldate end) / 2 as S,
    count(case when pba.spacetype = 'WC' then pb.ldate end) / 2 as WC,
    count(case when pba.spacetype = 'WK' then pb.ldate end) / 2 as WK
from 
    passbookingactivity as pba,
    passbooking as pb,
    passbookingleg as pbl,
    mappolygon as mp
where 
    pb.bookingid = pba.bookingid
    and pb.bookingid = pbl.bookingid
    and mp.polyid = pbl.addresspolygonid
    and pb.ldate = 20170403
    and pb.servicetypeid = 5
group by mp.name, mp.abbreviation

我想在mp.name中加入这些，然后按pr.runname，cp.firstname，mp.name

分组

Answer 1

虽然我没有把它弄得太漂亮，但这应该可以解决问题。 Lemme知道它是怎么回事！

select
a.runname
,a.firstname
, a.name
from 
(select
    pr.runName,
    cp.firstname,
    mp.name
from
    passrun as pr,
    passrunpoly as prp,
    mappolygon as mp,
    cmnemployee as ce,
    cmnperson as cp,
    passschedule as ps

where 
    pr.runid = prp.runid
    and prp.polyid = mp.polyid
    and pr.employeeid = ce.employeeid
    and ce.personid = cp.personid
    and pr.scheduleid = ps.scheduleid
    and ps.ldate = 20170403)a
inner join (
select 
    mp.name,
    count(distinct pbl.lat) as Stops,
    count(case when pba.spacetype = 'S' then pb.ldate end) / 2 as S,
    count(case when pba.spacetype = 'WC' then pb.ldate end) / 2 as WC,
    count(case when pba.spacetype = 'WK' then pb.ldate end) / 2 as WK
from 
    passbookingactivity as pba,
    passbooking as pb,
    passbookingleg as pbl,
    mappolygon as mp
where 
    pb.bookingid = pba.bookingid
    and pb.bookingid = pbl.bookingid
    and mp.polyid = pbl.addresspolygonid
    and pb.ldate = 20170403
    and pb.servicetypeid = 5
group by mp.name, mp.abbreviation)b
 on a.name = b.name
group by a.runname,a.firstname, a.name

Answer 2

您可以使用公用表表达式（我也更改了您的连接语法）：

with table_one as (
select
    pr.runName,
    cp.firstname,
    mp.name
from
    passrun as pr
    inner join passrunpoly as prp on pr.runid = prp.runid
    inner join mappolygon as mp on prp.polyid = mp.polyid
    inner join cmnemployee as ce on pr.employeeid = ce.employeeid
    inner join cmnperson as cp on ce.personid = cp.personid
    inner join passschedule as ps on pr.scheduleid = ps.scheduleid
where 
    ps.ldate = 20170403

), table_two as (

select 
    mp.name,
    count(distinct pbl.lat) as Stops,
    count(case when pba.spacetype = 'S' then pb.ldate end) / 2 as S,
    count(case when pba.spacetype = 'WC' then pb.ldate end) / 2 as WC,
    count(case when pba.spacetype = 'WK' then pb.ldate end) / 2 as WK
from 
    passbookingactivity as pba
    passbooking as pb on pb.bookingid = pba.bookingid
    passbookingleg as pbl on pb.bookingid = pbl.bookingid
    mappolygon as mp on mp.polyid = pbl.addresspolygonid
where 
    pb.ldate = 20170403
    and pb.servicetypeid = 5
group by mp.name, mp.abbreviation
)

select
    one.runname,
    one.name,
    one.firstname
from table_one one
    inner join table_two two on one.name = two.name
group by one.runname,one.name,one.firstname

SQL - 连接2个派生表

2 个答案: