一、介绍
这是关于mysql比较经典的练习题，在我们学习完mysql数据库以后，可以通过练习得到巩固
二、案例介绍
orderinfo 订单详情表（ orderid 订单id；userid 用户id； isPaid 是否支付；price 付款价格； paidTime 付款时间）
userinfo 用户信息表（userid 用户id； sex 用户性别； birth 用户出生日期）
三、题目要求及答案
1、统计不同月份的下单人数
# 首先明确一点，下单人数应该是已经支付的人数，并且用户可能不止一次下单，就需要去重,不同月份意味着分组
# 代码
"""
select year(paidTime), month(paidTime), count(distinct userid) as cons
from orderinfo
where isPaid="已支付" and paidTime<>'0000-00-00 00:00:00'
group by year(paidTime),month(paidTime);
"""
2、统计用户三月份的回购率和复购率
知识点： 复购率：当月购买了多次的用户占占当月用户的比例
回购率：上月购买的用户中有多少用户本月又再次购买
# 思路：三月份的复购率，也就是先统计出总用户人数，后面统计购买多次的人数，相比较就可以了
select count(ct),sum(if(ct>1,1,0)
(select userid,count(*) as ct
from orderinfo
where isPaid="已支付" and month(paidTime) = 3;
group by userid) a;
# 回购率（第一种解读：直接使用月份，month=3和4，分别算出数量，在join拿到数据，不做过多解读）
第二种解答：标准模板
select t1.dt, count(t1.userid),count(t2.userid),count(t2.userid)/count(t1.userid) as "回购率"
（select userid,date_format(paidtime,"%Y-%m-01") as dt
from orderinfo
where ispaid = "已支付"
group by userid,date_format(paidtime,"%Y-%m-01")） t1
left join
（select userid,date_format(paidtime,"%Y-%m-01") as dt
from orderinfo
where ispaid = "已支付"
group by userid,date_format(paidtime,"%Y-%m-01")） t2
on t1.userid = t2.userid
and t1.dt = date_sub(t2.dt,interval 1 month)
group by t1.dt
3、统计男女用户消费频次是否有差异
select
sex,
avg(cons) as avg_cons
from (select
a.userid,
sex,
count(1) as cons
from orderinfo a
inner join (select * from userinfo where sex<>'') b
on a.userid=b.userid
group by a.userid,sex) a
group by sex;
4、统计多次消费的用户，第一次和最后一次消费间隔是多少天
1.统计多次消费的用户,取时间（分组后取时间）
select userid,min(paidtime),max(paidmax),datediff(max(paidmax),min(paidtime)) as "时间间隔"
from userinfo
where ispaid = "已支付"
group by userid
having count(1) >1;
5、统计不同年龄段，用户的消费金额是否有差异
select
age,
avg(cons),
avg(prices)
from (select
a.userid,
age,
count(1) as cons,
sum(price) as prices
from orderinfo a
inner join (select
userid,
birth,
now(),
ceil(timestampdiff(year,birth,now())/10) as age
from userinfo
where birth>'1901-00-00') b
on a.userid=b.userid
group by a.userid,age) a
group by age;
6、统计消费的二八法则，消费的top20%用户，贡献了多少消费额
select
count(1) as cons,
sum(total_price) as all_price
from (
select
userid,
sum(price) as total_price
from orderinfo a
where isPaid="已支付"
group by userid
order by total_price desc
limit 17000) b ;