select p.name,e.employee_id,e.employee_name, (sum(num1+num2+num3)) as time,pra.wage/(sum(num1+num2+num3)) as bzgzl, pra.wage2 /(sum(num1+num2+num3))
from pra_manhour
pra left join base_project p on pra.project_id = p.id left join base_employee e on e.employee_pk=pra.employee_pk
group by e.employee_name,p.name,e.employee_id 差不多就是这么一个情况,每次都要/(sum(num1+num2+num3+......))很麻烦,能不能用一个变量代替啊?
from pra_manhour
pra left join base_project p on pra.project_id = p.id left join base_employee e on e.employee_pk=pra.employee_pk
group by e.employee_name,p.name,e.employee_id 差不多就是这么一个情况,每次都要/(sum(num1+num2+num3+......))很麻烦,能不能用一个变量代替啊?
from(
select p.name,e.employee_id,e.employee_name,sum(num1+num2+num3) as num,pra.wage,pra.wage2
from pra_manhour pra
left join base_project p on pra.project_id = p.id
left join base_employee e on e.employee_pk=pra.employee_pk
group by e.employee_name,p.name,e.employee_id,pra.wage,pra.wage2
) t