mysql - 选择按一段时间内花费的金额计算的前 10 条记录？

Question

目前，我正在努力检索按一段时间内支出金额计算的前 10 条记录。

MySQL 表:

create table `payment_holder` (
    `user_id` int (11),
    `amount` Decimal (6),
    `date_added` datetime 
); 

演示数据:

insert into `payment_holder` (`user_id`, `amount`, `date_added`) values('4','3.75','2012-03-15 00:41:39');
insert into `payment_holder` (`user_id`, `amount`, `date_added`) values('5','32.20','2012-03-15 00:42:10');
insert into `payment_holder` (`user_id`, `amount`, `date_added`) values('6','32.20','2012-03-15 00:42:58');
insert into `payment_holder` (`user_id`, `amount`, `date_added`) values('7','0.89','2012-03-15 00:48:05');
insert into `payment_holder` (`user_id`, `amount`, `date_added`) values('8','3.75','2012-03-15 00:50:54');
insert into `payment_holder` (`user_id`, `amount`, `date_added`) values('4','3.75','2012-03-15 00:41:39');
insert into `payment_holder` (`user_id`, `amount`, `date_added`) values('5','32.20','2012-03-15 00:42:10');
insert into `payment_holder` (`user_id`, `amount`, `date_added`) values('6','32.20','2012-03-15 00:42:58');
insert into `payment_holder` (`user_id`, `amount`, `date_added`) values('7','0.89','2012-03-15 00:48:05');
insert into `payment_holder` (`user_id`, `amount`, `date_added`) values('8','3.75','2012-03-15 00:50:54');

我想从这个示例中检索如下所示的结果:

user_id amount
------------------
6       64.40   
5       64.40   
4       7.5 
8       7.5 
7       1.78

简而言之，根据 date_added，2012 年哪个 user_id 的购买量最高？

Answer 1

你有没有试过这样的东西，它会返回所有年份的所有数据:

select user_id, sum(amount) Amount
from payment_holder
group by user_id
order by amount desc
limit 0, 10

参见 SQL Fiddle with Demo

但是如果你想按年份限制，你可以添加一个 WHERE 子句，它将 YEAR() 函数应用到 date_added领域:

select user_id, sum(amount) Amount
from payment_holder
where year(date_added) = 2012
group by user_id
order by amount desc
limit 0, 10