按城市计算收入超过 x 值和低于 x 值的员工人数

Question

表名为employee，列是city，salary 和 员工

目标：在单个查询中，按城市统计所有收入 >=1300 值小于

city	stipendio	employee
Milano	1200	employee1
Roma	1000	employee2
Milano	2000	employee3
Roma	900	employee4
Milano	1600	employee5

我单独做的查询：

#1

select
CITTA,
COUNT(*)
FROM
employee
WHERE city in ( 'Milano', 'Roma')
and salary>=1300 
GROUP BY city;

#2

select
city,
COUNT(*)
FROM
employee
WHERE city in ( 'Milano', 'Roma')
and salary<=1300
GROUP BY city;

Answer 1

使用条件聚合。也就是说，在聚合中包含 CASE 表达式。

SELECT
  city,
  SUM(CASE WHEN salary <= 1300 THEN 1 ELSE 0 END)  less_than_or_equal_to,
  SUM(CASE WHEN salary >  1300 THEN 1 ELSE 0 END)  greater_than
FROM
  employee
WHERE
  city IN ( 'Milano', 'Roma')
GROUP BY
  city;

Answer 2

试试这个。

select
city,
sum(case when salary<=1300 then 1 else 0 end) below,
sum(case when salary>1300 then 1 else 0 end) above


 from employee

WHERE city in ( 'Milano', 'Roma')
group by city