一、创建表，并将数据导入userinfo和orderinfo表中
1.导入数据：
第一步：建表
create table orderinfo(
orderid int primary key not null ,
userid int,
isPaid varchar(10),
price float,
paidTime varchar(30));
create table userinfo(
userid int primary key,
sex varchar(10),
birth date);
第二步：导数
load data local infile 'C:/Users/lan/Desktop/123/order_info_utf.csv' into table orderinfo fields terminated by ',';
load data local infile 'C:/Users/lan/Desktop/123/user_info_utf.csv' into table userinfo fields terminated by ',';
二、mysql数据分析案例
1.首先了解两个表的信息：
userinfo 用户信息表：
userid：主键，用户id
sex：用户性别
birth：用户生日
orderinfo 订单信息表
orderid：订单id
userid：用户id
ispaid：支付状态
price：支付价格
paidtime：支付时间
2.统计不同月份的下单人数
分析要点：不同月份，使用month函数获取订单日期中的月份，下单状态为已支付状态
select month(paidtime),count(distinct userid) from orderinfo where ispaid='已支付' group by month(paidtime);
3.统计用户三月份和四月份回购率和复购率
复购率：本月消费的人数有多少是重复购买的，即购买次数大于1次；
分步计算：
首先求出三月份存在消费的用户，以及每个用户的消费次数；（由于运行数据较大限制显示行数）
select userid,count(userid) as ct from orderinfo where ispaid='已支付' and month(paidtime)=3 group by userid limit 10;
如上表所示userid列为本月存在消费的用户，ct列为消费次数，统计ct列，count(ct)为计算本月存在消费总人数，通过if函数计算出消费次数大于1的消费者数量，即复购的消费者数，相除即为复购率；
select count(ct),count(if(ct>1,1,null)) as nt from
(select userid,count(userid) as ct from orderinfo where ispaid='已支付' and month(paidtime)=3 group by userid) t;
也可使用case when方法，结果相同；
select count(ct),count(case when ct>1 then 1 else null end) as nt from
(select userid,count(userid) as ct from orderinfo where ispaid='已支付' and month(paidtime)=3 group by userid) t;
回购率：三月份购买的人数，四月份依旧购买；
首先用userid和月份进行分组，计算用户在当月是否消费过；（由于运行数据较大限制显示行数）
select userid,date_format(paidtime,'%Y-%m-01') from orderinfo where ispaid='已支付' group by userid,date_format(paidtime,'%Y-%m-01') limit 10;
通过left join对表进行关联，关联出三月、四月都存在消费情况（由于运行数据较大限制显示行数）
如图，userid5 在3月和4月都存在消费，
select * from
(select userid,date_format(paidtime,'%Y-%m-01') as m 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 m from orderinfo where ispaid='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) limit 20;
对上表进行计数，计算出当月存在消费的情况和下月依然存在消费的情况，相除即为复购率；
select t1.m,count(t1.m),count(t2.m) from
(select userid,date_format(paidtime,'%Y-%m-01') as m 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 m from orderinfo where ispaid='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;
3.统计男女用户的消费频次是否有差异；
首先性别字段存在空值，先把空值过滤掉；
select * from userinfo where sex <>' ' limit 10;
将订单表与用户表相关联；
select * from orderinfo o
inner join
(select * from userinfo where sex<>' ') t
on o.userid=t.userid limit 20;
统计男、女的消费总次数；
select o.userid,sex,count(1) from orderinfo o
inner join
(select * from userinfo where sex<>' ') t
on o.userid=t.userid
group by o.userid,sex limit 20;
统计男女消费频次差异；
select sex,avg(ct) from (
select o.userid,sex,count(1) as ct from orderinfo o
inner join
(select * from userinfo where sex<>' ') t
on o.userid=t.userid
group by o.userid,sex) t2
group by sex;
4.统计多次消费的用户，第一次消费和最后一次消费的时间间隔
首先求出多次消费的用户；
select * from orderinfo where ispaid='已支付' group by userid having count(1)>1;
求出最大时间，和最小时间，时间间隔（求分组后的组内的最大值和最小值,按照userid 分组已经过滤掉了消费次数小于1的用户，所以至少存在两次消费情况，课选出最大时间，最小时间）
select userid,max(paidtime),min(paidtime) ,datediff(max(paidtime),min(paidtime)) as mt from orderinfo where ispaid='已支付' group by userid having count(1)>1 limit 20;
5.统计不同年龄段，消费是否有差异：
首先统计用户的年龄段；注意年龄计算函数，过滤无效数据；
select userid,ceil((year(now())-year(birth))/10) as ct from userinfo where birth>'1901-00-00' limit 10;
订单表和用户表关联，查看不同年龄段消费情况；
select * from orderinfo o
inner join
(select userid,ceil((year(now())-year(birth))/10) as ct from userinfo where birth>'1901-00-00') t
on o.userid=t.userid limit 10;
统计不同年龄段消费频次；
select o.userid,t.ct,count(t.ct) as mt from orderinfo o
inner join
(select userid,ceil((year(now())-year(birth))/10) as ct from userinfo where birth>'1901-00-00') t
on o.userid=t.userid group by o.userid,t.ct limit 10;
6.统计消费二八法则，消费20%的用户，贡献了多少额度；
首先计算用户的消费总金额；
select userid,sum(price) from orderinfo where ispaid='已支付' group by userid limit 10;
select userid,sum(price) as total from orderinfo where ispaid='已支付' group by userid limit 10;
计算出前20%的用户；
select count(userid) *0.2,sum(total) from (
select userid,sum(price) as total from orderinfo where ispaid='已支付' group by userid order by total) t;
由于这里计算出的结果是前17000行，所以将内层查询条件限制为17000；
select count(userid) ,sum(total) from (
select userid,sum(price) as total from orderinfo where ispaid='已支付' group by userid order by total desc limit 17000) t;