Question

我有一个包含id（int）和sell_date（datetime）的表。这些是客户id和销售日期。我必须得到客户的数量，他们每个月至少买2次，按DATE_FORMAT(sell_date,'%Y-%m') AS period分组。

例如：我的客户端id = 1。他在01 - 2014年买了一次东西，在02-2014买了2次，在03-2014买了3次。所以我想得到这个：

period  |repeated_buyers

2014-01 | (none)
2014-02 | 1
2014-03 | 1

我不如SQL。非常感谢你提前！

Answer 1

你快到了。您需要在日期<{1}}上进行<{1}}

group by

Answer 2

试试这个：

select DATE_FORMAT(sell_date,'%Y-%m') AS period,COUNT(ID) as repeated_buyers
from MyTable 
GROUP BY ID,DATE_FORMAT(sell_date,'%Y-%m')
having COUNT(ID)>=2

Answer 3

啊，我做到了！

    SELECT `period`, 
count(`id`) as `accounts`
    FROM (SELECT 
DATE_FORMAT(`sell_date`,'%Y-%m') AS `period`, 
`id` FROM `terst` GROUP BY `period`, `id` HAVING COUNT(`id`)>1 )A
    GROUP BY `period`

Answer 4

尝试下面的sql查询，

select id,period,repeated_buyers from(select DATE_FORMAT(sell_date,'%Y-%m') AS period,id,
Count(*) as repeated_buyers
from TABLE1
group by  DATE_FORMAT(sell_date,'%Y-%m'),id) where repeated_buyers>=2;

SQL查询以获取所有重复的条目

4 个答案: