排名在postgres总数

Question

我正在写一个SQL脚本，我想要得到每个销售人员的约会总数，然后也得到多少他从其他销售人员排名。例如销售员x有5个预约，他对10个推销员中的4个进行评分。

**expected results**: 
Salesperson x 5 4/10 
Salesperson D 6 5/10 
Salesperson s 8 7/10 

Answer 1

使用rank()

with sales as 
(
select Salesperson, count(appointment) appointments 
from SalesTable 
group by Salesperson 
) 
select sales.*, rank() over (order by appointments desc) as salesrank 
from sales 

Answer 2

您好感谢您的答复。我尝试了这种方式它的工作原理：

select id,sales_person,"Appointment/Day",rank_for_the_day,"Appointment/Week",rank_for_the_week,"Appointment/Month", 
rank_for_the_month,"Appointment/year",rank_for_the_year 
from(
select supplied_id,salesperson,sum(case when appointment_date::date=current_date then 1 else 0 end)"Appointment/Day", 
    rank() over (order by sum(case when appointment_date::date=current_date then 1 else 0 end) desc)||'/'|| 
    (select sum(case when appointment_date::date=current_date then 1 else 0 end) from match where date_part('year', appointment_date)=2017 
    and appointment_date is not null and date_part('day',appointment_date)=date_part('day',current_date)) rank_for_the_day, 

    sum(case when appointment_date::date between current_date-7 and current_date then 1 else 0 end)"Appointment/Week", 
    rank() over (order by sum(case when appointment_date::date between current_date-7 and current_date then 1 else 0 end) desc)||'/'|| 
      (select sum(case when appointment_date::date between current_date-7 and current_date then 1 else 0 end) 
      from match m where date_part('year', appointment_date)=2017 and appointment_date is not null 
      and date_part('week',appointment_date)=date_part('week',current_date)) rank_for_the_week, 

    sum(case when date_part('month',appointment_date)=date_part('month',current_date) then 1 else 0 end)"Appointment/Month", 
     rank() over (order by sum(case when date_part('month',appointment_date)=date_part('month',current_date) then 1 else 0 end) desc)||'/'|| 
      (select sum(case when date_part('month',appointment_date)=date_part('month',current_date) then 1 else 0 end) 
      from match m where date_part('year', appointment_date)=2017 and appointment_date is not null 
      and date_part('month',appointment_date)=date_part('month',current_date)) rank_for_the_month, 

     sum(case when date_part('year',appointment_date)=date_part('year',current_date) then 1 else 0 end)"Appointment/year", 
    rank() over (order by sum(case when date_part('year',appointment_date)=date_part('year',current_date) then 1 else 0 end) desc)||'/'|| 
    (select sum(case when date_part('year',appointment_date)=date_part('year',current_date) then 1 else 0 end) 
      from match m where date_part('year', appointment_date)=2017 and appointment_date is not null 
      and date_part('year',appointment_date)=date_part('year',current_date)) rank_for_the_year 
from salespersontable 
where date_part('year', appointment_date)=2017 and appointment_date is not null 
group by id,salesperson 
)x order by 6 desc 

不过，我将不胜感激写这个查询，以尽量减少资源消耗的有效途径。