日期:2014-05-18 浏览次数:20701 次
select
h.hospitalname,i.illTypeName,
isnull(p.[有取药记录的人数],0) as [有取药记录的人数],
isnull(p.[没有取药记录的人数],0) as [没有取药记录的人数]
from illType i
Cross join hospital h
left join
(select illtypeID1,p.hospitalid,
sum(case when P2.patientID is not null then 1 else 0 end) as [有取药记录的人数],
sum(case when P2.patientID is null then 1 else 0 end) as [没有取药记录的人数]
from patient_info as p left join payInfo as P2 on P2.patientID=p.patientID
group by illtypeID1,p.hospitalid
)p on p.illtypeID1 = i.illtypeID and h.hospitalid = p.hospitalid