Question

我必须编写一个查询来计算每个客户购物之间的平均天数（不使用子查询）。

    create table data {
customer varchar(20) not null,
bought date not null,
primary key (customer,bought)
}

例如，

insert into data (customer,bought)
values (‘John Smith’, date ‘2011-02-01’),
(‘Alice Cooper’, date ‘2011-02-01’),
(‘Bob Baker’, date ‘2011-02-01’),
(‘John Smith’, date ‘2011-02-02’),
(‘Bob Baker’, date ‘2011-02-02’),
(‘Bob Baker’, date ‘2011-02-03’),
(‘Bob Baker’, date ‘2011-02-04’),
(‘Bob Baker’, date ‘2011-02-05’),
(‘Bob Baker’, date ‘2011-02-06’),
(‘Bob Baker’, date ‘2011-02-07’),
(‘John Smith’, date ‘2011-02-07’),
(‘Alice Cooper’, date ‘2011-02-08’);

应该返回John Smith等待1天然后5天，所以他的平均值是3天。 Alice Cooper（！）等了7天所以她平均是7.鲍勃贝克是每日跑步者，所以他的平均值是1。

我做过类似的事情

select distinct customer, avg (bought) as average from data;

但它不起作用。

非常感谢任何帮助。

Answer 1

您必须在epoch秒内转换时间戳才能使用avg聚合函数：

SELECT
    customer,
    timestamp without time zone '1970-01-01' + cast(
       avg(EXTRACT(EPOCH FROM bought::timestamp)
    )::text as interval) 
FROM data
GROUP BY customer;

Answer 2

PostgreSQL版本的链接答案

select customer, (max(bought) - min(bought)) / (count(bought)-1)
from data
group by customer;

Answer 3

这在类似的情况下对我有用：

SELECT customer, avg(AGE(now(),bought)) as waited
FROM data 
GROUP BY customer

Answer 4

选择max-min / count不是平均值。尝试：

 timestamp 'epoch' + avg(date_part(epoch, date_column))  * interval '1 second'

Answer 5

您可能希望使用分组声明

select 
   customer, 
   datediff(D, min(bought), max(bought)) / count(bought) as average
from data
group by customer

如果选择列表中有聚合函数，则必须对非聚合成员的其他字段使用分组。

这是在SQL Server上测试的，语法可能与Postgresql不同，后者我没有使用。

Answer 6

SELECT customer, AVG(bought) AS average FROM data GROUP BY customer

Sql：日期的平均值

6 个答案: