数据库练习题：使用mysql分析销售数据

销售数据文件链接(失效了可以在评论区艾特)：
https://pan.baidu.com/s/1-thB0rrXYmkSoey30lwYLg 提取码：kmr8

使用mysql完成上面的6个问题。首先导入本地csv文件到mysql，有两种方式。第一种，直接导入，如下图所示：

注意：csv表要有表头，不然会导入不进去。
但是这种方式导入数据很慢，采取第二种方式，用代码导入很快

导入两个数据表order_info和user_info，其中order_info包含的信息如下：

user_info包含的信息如下：

问题1：统计不同月份的下单人数

select month(paidtime),count(distinct userid)
from order_info
where status !='未支付'
group by month(paidtime);

问题2.统计用户三月份的回购率和复购率

select userid,count(userid)
from order_info
where status ='已支付'
and month(paidtime)=3
group by userid;

先计算出3月份每个用户的购买次数，复购率就等于购买次数大于1的人数除以购买的总人数

select count(ct) '购买人数',
count(if(ct>1,1,null)) '回购人数' ,
count(if(ct>1,1,null))/count(ct) '复购率'
from(
    select userid,Count(userid) as ct from order_info
    where status = '已支付'
    and month(paidTime) = 3
    group by userid
    order by userid) t;

结果如下

对于回购率，就是计算一段时间间隔内用户再次购买的概率

select t1.m '月份',
count(t1.m) '消费总数' ,
count(t2.m) '回购数' ,
count(t2.m)/count(t1.m) '回购率'
from (
select userid,date_format(paidtime,'%Y-%m-01') as m
from order_info
where status = '已支付'
group by userid,date_format(paidtime,'%Y-%m-01') 
) t1
left join (
select userid,date_format(paidtime,'%Y-%m-01') as m
from order_info
where status = '已支付'
group by userid,date_format(paidtime,'%Y-%m-01') 
) t2
on t1.userid=t2.userid
and t1.m = date_sub(t2.m,interval 1 month)
group by t1.m;

问题3.统计男女用户的消费频次是否有差异

select sex,avg(ct) from(
    select o.userid,sex,count(o.userid)as ct from order_info o
    inner join
        (select * from user_info
        where sex != '') t
    on o.userid = t.userid
    group by userid,sex
    order by userid)t2
group by sex;

从结果可以看出并无差异

问题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?

select userid,max(paidtime),min(paidtime),datediff(max(paidtime),min(paidtime))
from order_info
where status = '已支付'
group by userid 
having count(1)>1;

问题5.统计不同年龄段，用户的消费金额是否有差异？

select o.userid,age,sum(price),count(o.userid) as ct
from order_info o
left join (
select userid,ceil((year(now())-year(birth))/10) as age 
from user_info 
where birth>1901-00-00) t
on o.userid=t.userid
where status = '已支付'
group by age;

问题6.统计消费的二八法则，消费的top20%用户，贡献了多少额度

首先计算总的消费额度

select count(userid),sum(total) 
from (
select userid,sum(price) as total 
from order_info
where status = '已支付'
group by userid
order by total DESC
) as t;

结果如下图

然后计算消费的top20%

select count(userid)*0.2,sum(total) 
from (
select userid,sum(price) as total 
from order_info
where status = '已支付'
group by userid
order by total DESC
) as t;

select count(userid),sum(total) 
from (
select userid,sum(price) as total 
from order_info
where status = '已支付'
group by userid
order by total DESC
limit 17129
) as t;

然后用前20%的消费额度除以上面得到的总额度就得到了top20%的消费额度占比。

最近找实习找的挺惆怅的，希望可以快点找到好实习，找到想去的公司，就不用天天这么丧了
好想有一个数据分析行业相关的朋友给我指点指点…双非硕真的这么不行么…还是我不行…