用MySQL分析网络销售案例

用MySQL分析网络销售案例

数据来源于某网站销售统计

1. 网络订单数据
2. 用户信息

点击获取数据 提取码：3k6i

分析步骤

0、数据导入

1、不同月份的下单人数

2、用户三月份的回购率和复购率

3、统计男女用户的消费频次

4、统计多次消费用户，分析第一次和最后一次的消费间隔

5、统计不同年龄段用户的消费金额差异

6、统计消费的二八法则：消费top20%的用户贡献了多少消费额度

0 数据导入

首先需要先创建对应的数据库和相应的表

1. 创建orderinfo 表

2. 创建userinfo表

3. 登录mysql导入相应的数据

load data local infile "file" into table dbname.tablename ...

# 登录
mysql --local-infile -uroot -p
# 导入数据orderinfo
load data local  infile 'F:\BaiduNetdiskDownload\SQL\order_info_utf.csv' into table data.orderinfo fields terminated by ',';
# 导入数据userinfo
load data local  infile 'F:\BaiduNetdiskDownload\SQL\user_info_utf.csv' into table data.userinfo fields terminated by ',';

4. 观察数据，对时间进行处理 ; 更新字符串为日期格式

update orderinfo set paidtime=replace(paidtime,'/','-') where paidtime is not null

update orderinfo set paidtime=str_to_date(paidtime,'%Y-%m-%d %H:%i') where paidtime is not null

5查看数据

1 不同月份的下单人数

思路 ：按月份进行分组，对用户进行去重统计

select month(paidTime) as dtmonth, 
count(distinct userId) as count_users
from orderinfo
where isPaid = '已支付'
group by month(paidTime)

2 用户三月份的回购率和复购率

1. 复购率 ： 自然月内，购买多次的用户占比

• 首先先找出已支付中3月份的用户id和对应次数，按用户分组
• 然后再嵌套一层，复购率：购买次数大于1/ 总购买次数

select count(ct),count(if(ct>1,1,null)) from(
	select userID,Count(userId) as ct from orderinfo
	where isPaid = '已支付'
	and month(paidTime) = 3
	group by userId
	order by userId) t

复购率： 16916 / 54799 = 0.308

2. 回购率： 曾经购买过的用户在某一时期内再次购买的占比

首先先查询已支付userId ,和 支付月份的统计

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')

然后使用date_sub函数，将表关联，筛选出本月的消费的userID，和下月的回购userID，即可计算出回购率

select t1.m,count(t1.m) as 消费总数,
 count(t2.m) as 复购率,
 count(t2.m)/ count(t1.m) as 回购率 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

• userinfo因为性别有空值，需要筛选出t orderinfo 再和表t连接 统计出用户男女消费次数

select o.userId,sex,count(o.userId)as ct from orderinfo o
	inner join
		(select * from userinfo
		where sex != '') t
	on o.userId = t.userId
	group by userId,sex
	order by userId

• 根据上表，在进行子查询，统计出男性消费频次

select sex,avg(ct) from(
	select o.userId,sex,count(o.userId)as ct from orderinfo o
	inner join
		(select * from userinfo
		where sex != '') t
	on o.userId = t.userId
	group by userId,sex
	order by userId)t2
group by sex

4 统计多次消费用户，分析第一次和最后一次的消费间隔

• 首先把多次消费的用户,和相应第一次最后一次消费时间提取出来
• 然后使用datediff 计算时间间隔，以天为单位

select userId,max(paidTime),min(paidTime),datediff(max(paidTime),min(paidTime)) from data.orderinfo
where isPaid = '已支付'
group by userId having count(1) > 1
order by userId

5 统计不同年龄段用户的消费金额差异

通过表联结，给用户划分不同的年龄段，以10年为基准，过滤出生日期为1900-00-00的异常值,筛选出用户消费频次和消费金额

select o.userId,age,price,count(o.userId)as ct from orderinfo o
inner join (
	select userId, ceil((year(now()) - year(birth))/10) as age
			from userinfo
			where birth > 1901-00-00) t
on o.userId = t.userId
where isPaid = '已支付'
group by userId
order by userId

统计出年龄段的消费频次和消费金额

select t2.age,avg(ct),avg(price) from (
	select o.userId,age,price,count(o.userId)as ct from orderinfo o 
	inner join(
		select userId, ceil((year(now()) - year(birth))/10) as age
		from userinfo
		where birth > 1901-00-00)t
	on o.userId = t.userId
    where ispaid = '已支付'
	group by userId, age) t2
group by age
order by age

• ceil : 向上取整

6 统计消费的二八法则：消费top20%的用户贡献了多少消费额度

按照用户消费总额排序

select userId,sum(price) as total from orderinfo o
where isPaid = '已支付'
group by userId
order by total desc

查看总用户数和总金额

select count(userId),sum(total) from (
	select userId,sum(price) as total from orderinfo o
	where isPaid = '已支付'
	group by userId
	order by total desc) as t

查看前20%的用户数量有多少

select count(userId)*0.2,sum(total) from (
	select userId,sum(price) as total from orderinfo o
		where isPaid = '已支付'
		group by userId
		order by total desc)as t

limit限制前17000用户

select count(userId),sum(total) from (
select userId,sum(price) as total from orderinfo o
where isPaid = '已支付'
group by userId
order by total desc
limit 17129) t

top20%用户的消费总额占比情况:top20%用户的消费总额/所有用户的消费总额=73.93%
top20%的用户贡献了73.93%消费额度。