Question

我有2张桌子。我有一个sql，可以显示时间，日期，employee_id，并计算重复ID。现在我想允许系统可以统计所有以字母开头的名字＆＃39; L＆＃39;并显示每天的所有员工名字的总和以字母＆＃39; L＆＃39;

select distinct a.employee_id,b.name,a.type,a.date,a.time,count(*) 
from deal_records a, staff_table b 
where a.employee_id = b.employee_id && a.date>='2017-1-1' 
  and a.date <'2017-2-1'&& b.name like "L%" 
group by a.employee_id;

The format should be looks like below

Answer 1

在单个查询中同时包含select distinct和group by的气味很多，请适当地扩展group by子句。

关于回答问题，请将count(*)替换为sum(case when b.name like 'L%' then 1 end)。

Answer 2

select a.employee_id,b.name,a.type,a.date,a.time,
sum( case when b.name like "L%" 1 else 0 end) as all_count
from deal_records a, staff_table b 
where a.employee_id = b.employee_id 
&& trunc(a.date) between to_date('01.01.2017', 'DD.MM.YYYY') and to_date('01.02.2017', 'DD.MM.YYYY')
group by a.employee_id
order by trunc(a.date)

sql如何在一个月内计算总和

2 个答案: