case..then..oracle函数的研究.......

标签：

SELECT a.areaname as 二级地名 FROM houses h,houses h2,
areas a WHERE h.areaid=a.areasid AND h2.areaid=a.areasid AND h.finalassessmentid=1 AND
h.houseslevel=2 AND h2.houseslevel=1 AND a.areasid!=a.parentareasid AND a.companysid=2;

select * from areas
select count(h.houseslevel) as 总数,count(h1.houseslevel) as 最新累计成盘,count(h2.houseslevel) as 最新累计必交盘 from houses h,houses h1,houses h2 where h.houseslevel in (1,2) and h1.houseslevel=1 and h2.houseslevel=2;

select count(h.houseslevel) as 总数 from houses h where h.houseslevel in (1,2)
union all
select count(h.houseslevel) as 成盘 from houses h where h.houseslevel=1
union all
select count(h.houseslevel) as 成盘 from houses h where h.houseslevel=2;

select
count(h.houseslevel) as 总数,
count(case when h.houseslevel=1 then h.housesid end ) 成盘 ,
count(case when h.houseslevel=2 then h.housesid end ) 比较盘
from houses h,areas a where h.areaid=a.areasid and h.houseslevel in (1,2)

SELECT a.areaname AS 二级地名,
count(a.houseId) AS 最新累计有效待售数,
count(h.houseslevel) AS 最新累计必卖盘,count(h2.houseslevel) AS 最新累计成盘 FROM houses h,houses h2,
areas a WHERE h.areaid=a.areasid AND h2.areaid=a.areasid AND h.finalassessmentid=1 AND h.houseslevel=2 AND h2.houseslevel=1 AND a.areasid!=a.parentareasid AND a.companysid=2;

SELECT a.areaname AS 二级地名 FROM houses h,areas a WHERE h.areaid=a.areasid AND a.areasid!=a.parentareasid

SELECT a.areaname AS 二级地名 FROM areas a where a.areaname=

select a.areaname AS 二级地名 from areas a
where a.areaname in (select b.areaname from areas b group by b.areaname having count(b.areaname) > 1)

select b.areaname as 二级地名,count(b.areaname) as 重复数据量 from areas b group by b.areaname having count(b.areaname) > 1

select h.houseslevel from houses h where h.houseslevel=1;

select a.areaname AS 二级地名 from houses h,areas a where h.areaid=h.areaid and h.houseslevel=1 and a.areasid!=a.parentareasid AND a.companysid=2 and h.finalassessmentid=1;

select a.areaname AS 二级地名,
count(h.houseslevel) as 总数,
count(case when h.houseslevel=1 then h.housesid end ) 成盘 ,
count(case when h.houseslevel=2 then h.housesid end ) 比较盘
from houses h,areas a where h.areaid=a.areasid and h.houseslevel in (1,2) and a.areasid!=a.parentareasid AND a.companysid=2 and h.finalassessmentid=1 and flag1=1 group by a.areaname;

case..then..oracle函数的研究.......

标签：

原文地址：http://www.cnblogs.com/shunzdd/p/5585985.html