Как получить сумму значений столбца «temp» из этого запроса?

Question

введите описание изображения здесь Как получить сумму временного столбца:

select Split_Part("EmpName",'_',1),
       "EmployeeID",
       sum("TotalDays"::decimal) as "temp" 
from "Leave_Log" 
where date("StartDate") between '2018-01-01' and '2018-12-31' 
  and "Staus"='Approved' 
  and "EmployeeID" in (Select "UserName" 
                      from  "Master_Employees" 
                      where "Status"='Y')
group by "EmpName","EmployeeID" 
order by "EmpName"

Answer 1

Если вы хотите получить столбец с total sum of temp, вам следует использовать Функции окна

select Split_Part("EmpName",'_',1) splited_name,
       "EmployeeID",
       sum("TotalDays"::decimal) as "temp",
       sum(sum("TotalDays"::decimal)) OVER () as total

Answer 2

Неправильно выглядит GROUP BY:

select Split_Part("EmpName",'_',1) splited_name,
       "EmployeeID",
       sum("TotalDays"::decimal) as "temp" 
from "Leave_Log" 
where date("StartDate") between '2018-01-01' and '2018-12-31' 
  and "Staus"='Approved' 
  and "EmployeeID" in (Select "UserName" 
                      from  "Master_Employees" 
                      where "Status"='Y')
group by splited_name,"EmployeeID" 
order by splited_name

например: если у вас есть таблица

CREATE TABLE public.users
(
  id integer NOT NULL DEFAULT nextval('users_user_id_seq'::regclass),
  email character varying(100) NOT NULL,
  salary integer,
  CONSTRAINT users_pkey PRIMARY KEY (id),
  CONSTRAINT users_email_key UNIQUE (email)
)

, тогда запросы

SELECT Split_Part(email, '@', 2) split_email, sum(salary)
  FROM users
  GROUP BY email
  ORDER BY email

AND

SELECT Split_Part(email, '@', 2) split_email, sum(salary)
  FROM users
  GROUP BY split_email
  ORDER BY split_email

даст вам другой ответ