SQL 收入最低的地点计数

Question

我正在使用在线编辑器来解决查询

https://www.jdoodle.com/execute-sql-online/

有如下SQL表

create table CITIES(id int, name String);
create table USERS(id int, city_id String, name String, email);
create table RIDES(id int, user_id int, distance int, fare int);

INSERT INTO CITIES (id, name) VALUES
(1,"Cooktown"),
(2,"South Suzanne");

INSERT INTO USERS (id, city_id, name, email) VALUES
(1,2,"a","email"),
(2,2,"b","email"),
(3,1,"c","email"),
(4,1,"d","email"),
(5,1,"e","email"),
(5,1,"f","email");

INSERT INTO RIDES (id, user_id, distance, fare) VALUES
(1,1,21,200),
(2,3,6,55),
(3,2,30,230),
(4,2,16,125),
(5,2,11,110),
(6,6,30,285),
(7,3,18,170),
(8,1,6,50),
(9,2,4,40),
(10,1,10,90),
(11,5,11,95),
(12,5,16,140),
(13,4,24,220),
(14,6,17,160),
(15,2,23,205),
(16,2,11,90),
(17,6,5,50),
(18,3,19,180),
(19,5,22,205),
(20,4,6,60);

输出应该是:

南苏珊娜 1050

库克敦 1710

我尝试了以下 SQL 查询，但不知道如何计数

select c.name, SUM(r.fare) from CITIES c
  left join USERS u on c.id=u.city_id 
  left join RIDES r on u.id=r.user_id

获取以下输出

库克敦|2705

任何帮助将不胜感激

Answer 1

看起来您缺少 GROUP BY:

SELECT c.name, SUM(r.fare) [SumOfFare]
FROM CITIES c
  LEFT JOIN USERS u ON c.id=u.city_id 
  LEFT JOIN RIDES r ON u.id=r.user_id
GROUP BY c.name
ORDER BY SUM(r.fare) ASC;

此外，您可能还想添加一个 ORDER BY 子句，这样您就可以在顶部看到收入最低的位置。