从表中选择用户的最大总和，作为最大数量的用户？

Question

我有一张桌子

create table mock_sales
(
    first_name character varying,
    last_name character varying,
    amount integer
);

insert into mock_sales(first_name, last_name, amount) 
values('ted','mosby', 100),
('lily', 'aldrin', 400),
('ted', 'mosby', 350),
('barney', 'Stinson',180)

所需的输出

金额最高的人

ted mosby // As ted mosby sum = 450 (100 + 350)，最大

我试过了

Select first_name, last_name from mock_sales group by first_name, last_name where amount in (
select max(amount) from t
(select sum(amount) as amount from mock_sales as t group by first_name, last_name)

or 

select t.first_name, t.last_name from mock_sales where max(amount) == t.amount and t.amount in (
Select first-name, last_name, sum(amount) as amount from mock_sales as t group by first_name, last_name)

但他们都给出了语法错误。任何帮助将不胜感激。

连接 2 个查询的结果时遇到问题。

Answer 1

既然你想要总和而不是最大值，那么试试这个：

SELECT first_name,last_name, SUM(amount) AS amount FROM mock_sales 
 GROUP BY first_name,last_name ORDER BY amount DESC LIMIT 1;

Answer 2

您可以只对具有相同名字/姓氏的记录进行分组，对结果进行排序并仅保留第一行：

select first_name, last_name, sum(amount) total_amount
from mock_sales
group by first_name, last_name
order by total_amount desc
limit 1

如果你想允许平局，那就有点不同了。在 Postgres 中，您可以使用窗口函数：

select *
from (
    select first_name, last_name, sum(amount) total_amount,
        rank() over(order by sum(amount) desc) rn
    from mock_sales
    group by first_name, last_name
) t
where rn = 1