日期:2014-05-18  浏览次数:20642 次

怎样将这两个查询结果 full jion起来?(当两个查询结果中的 UID和ProjectID都相等时 则进行合并)
第一个查询结果
SQL code

SELECT UID ,ProjectID,
      sum([LocalValidCount]) as 本地预约量
      ,sum([NoLocalValidCount]) as 外地预约量
  FROM DialogueQuantity
group by
  UID,ProjectID



第二个查询结果
SQL code

select
   UID,ProjectID,
   count(1) as 患者总数,
   sum(case IsNative when 'true' then 1 else 0 end) as '本地患者数量',
   sum(case IsNative when 'false' then 1 else 0 end) as '外地患者数量',
   sum(case DiseaseName when '感冒' then 1 else 0 end) as '感冒',
   sum(case DiseaseName when '发烧' then 1 else 0 end) as '发烧',
   sum(case DiseaseName when '阑尾炎' then 1 else 0 end) as '阑尾炎'




注:
要full jion起来
合并条件是:两个查询结果中的 UID和ProjectID都相等时 进行合并

------解决方案--------------------
SQL code
select * from
(
SELECT UID ,ProjectID,
      sum([LocalValidCount]) as 本地预约量
      ,sum([NoLocalValidCount]) as 外地预约量
  FROM DialogueQuantity
group by
  UID,ProjectI
) as a
full join 
(
select
   UID,ProjectID,
   count(1) as 患者总数,
   sum(case IsNative when 'true' then 1 else 0 end) as '本地患者数量',
   sum(case IsNative when 'false' then 1 else 0 end) as '外地患者数量',
   sum(case DiseaseName when '感冒' then 1 else 0 end) as '感冒',
   sum(case DiseaseName when '发烧' then 1 else 0 end) as '发烧',
   sum(case DiseaseName when '阑尾炎' then 1 else 0 end) as '阑尾炎'
) as b
on a.UID=b.UID and a.ProjectID=b.ProjectID

------解决方案--------------------
直接写 select * from
(
SELECT UID ,ProjectID,
sum([LocalValidCount]) as 本地预约量
,sum([NoLocalValidCount]) as 外地预约量
FROM DialogueQuantity
group by
UID,ProjectI
) as a
full join 
(
select
UID,ProjectID,
count(1) as 患者总数,
sum(case IsNative when 'true' then 1 else 0 end) as '本地患者数量',
sum(case IsNative when 'false' then 1 else 0 end) as '外地患者数量',
sum(case DiseaseName when '感冒' then 1 else 0 end) as '感冒',
sum(case DiseaseName when '发烧' then 1 else 0 end) as '发烧',
sum(case DiseaseName when '阑尾炎' then 1 else 0 end) as '阑尾炎'
) as b
on a.UID=b.UID and a.ProjectID=b.ProjectID
------解决方案--------------------
SQL code

-- 方法1
select * 
from ([子查询1]) a
full join ([子查询2]) b

-- 方法2
with 
a as
([子查询1]),
b as
([子查询2])
select *
from a full join b

------解决方案--------------------
SQL code

select * from (
SELECT UID ,ProjectID,
      sum([LocalValidCount]) as 本地预约量
      ,sum([NoLocalValidCount]) as 外地预约量
  FROM DialogueQuantity
group by
  UID,ProjectID)a full join (
select
   UID,ProjectID,
   count(1) as 患者总数,
   sum(case IsNative when 'true' then 1 else 0 end) as '本地患者数量',
   sum(case IsNative when 'false' then 1 else 0 end) as '外地患者数量',
   sum(case DiseaseName when '感冒' then 1 else 0 end) as '感冒',
   sum(case DiseaseName when '发烧' then 1 else 0 end) as '发烧',
   sum(case DiseaseName when '阑尾炎' then 1 else 0 end) as '阑尾炎'
) b on a.UID=b.UID and a.ProjectID and b.ProjectID