Question

目前，我正在努力检索在一段时间内按金额计算的前10名记录。

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');

我想从此示例中检索如下结果：

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

简而言之，2012年基于user_id的{{1}}购买量最高？

Answer 1

你有没有尝试过这样的东西，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

请参阅SQL Fiddle with Demo

但是，如果您希望按年限制，可以添加WHERE子句，将YEAR()函数应用于date_added字段：

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

选择按一段时间内花费的金额计算的前10个记录？

1 个答案: