我写了一个查询,其中一列是一个月。从那里我必须得到最小月份、最大月份和中位数月份。以下是我的查询。
select ext.employee,
pl.fromdate,
ext.FULL_INC as full_inc,
prevExt.FULL_INC as prevInc,
(extract(year from age (pl.fromdate))*12 +extract(month from age (pl.fromdate))) as month,
case
when prevExt.FULL_INC is not null then (ext.FULL_INC -coalesce(prevExt.FULL_INC,0))
else 0
end as difference,
(case when prevExt.FULL_INC is not null then (ext.FULL_INC - prevExt.FULL_INC) / prevExt.FULL_INC*100 else 0 end) as percent
from pl_payroll pl
inner join pl_extpayfile ext
on pl.cid = ext.payrollid
and ext.FULL_INC is not null
left outer join pl_extpayfile prevExt
on prevExt.employee = ext.employee
and prevExt.cid = (select max (cid) from pl_extpayfile
where employee = prevExt.employee
and payrollid = (
select max(p.cid)
from pl_extpayfile,
pl_payroll p
where p.cid = payrollid
and pl_extpayfile.employee = prevExt.employee
and p.fromdate < pl.fromdate
))
and coalesce(prevExt.FULL_INC, 0) > 0
where ext.employee = 17
and (exists (
select employee
from pl_extpayfile preext
where preext.employee = ext.employee
and preext.FULL_INC <> ext.FULL_INC
and payrollid in (
select cid
from pl_payroll
where cid = (
select max(p.cid)
from pl_extpayfile,
pl_payroll p
where p.cid = payrollid
and pl_extpayfile.employee = preext.employee
and p.fromdate < pl.fromdate
)
)
)
or not exists (
select employee
from pl_extpayfile fext,
pl_payroll p
where fext.employee = ext.employee
and p.cid = fext.payrollid
and p.fromdate < pl.fromdate
and fext.FULL_INC > 0
)
)
order by employee,
ext.payrollid desc
如果不可能,是否可以获得最大月份和最小月份?
最佳答案
要在 PostgreSQL 中计算中位数,只需取 50% 的百分位数(无需添加额外的函数或任何东西):
SELECT PERCENTILE_CONT(0.5) WITHIN GROUP(ORDER BY x) FROM t;
关于postgresql - 如何从 postgresql 中的查询中获取最小值、中值和最大值?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/12067656/