mysql视频练习题

mysql视频练习题

2个表：

• order_info_utf.csv
• user_info_utf.csv

导入到mysql数据库。

题目：

1.统计不同月份下单的人数。

⚠️这里的人数是指共有多少名自然人下单，不是指“人次”。所以count()内加上distinct，去重复。

SELECT month(paidTime), count(distinct userId) FROM test1.orderinfo
where isPaid = "已支付"
group by month(paidTime);

+-----------------+------------------------+
| month(paidTime) | count(distinct userId) |
+-----------------+------------------------+
|               3 |                  54799 |
|               4 |                  43967 |
|               5 |                      6 |
+-----------------+------------------------+
3 rows in set (1.23 sec)

---

2.统计用户3月的回购率和复购率

2.1复购率：本月消费1次以上的人数/本月消费的总人数。

    SELECT userid, count(userid) as ct FROM test1.orderinfo
    where isPaid = "已支付"
    and month(paidTime) = 3
    group by userid having ct > 1;

得到一个列表：本月消费1次以上的人的id和他的消费次数的集合。

这里使用了having筛选出消费1次以上的人的id。

更好的方法是，查询结果显示2个列，分别储存“本月消费1次以上的人数”， “本月消费的总人数”：

所以，需要外面再加一层，并去掉having：

select count(if(ct > 1, 1, null)), count(1) from (
    SELECT userid, count(userid) as ct FROM test1.orderinfo
    where isPaid = "已支付"
    and month(paidTime) = 3
    group by userid) as t

得到2个数字16916和54799，所以复购率是：30.87%

2.2回购率：3月消费的人，又在4月进行了消费。这是一次回购：x。x除以3月消费的人数= 3月回购率。

3月消费的人数: 54799人

select count(1) from (
    select userid from test1.orderinfo
    where isPaid = "已支付"
    and month(paidTime) = 3
    group by userid) as user

3月消费的人，又在4月进行了消费的人数： 13119人

select count(1) from ( 
    select userid from test1.orderinfo
    where isPaid = "已支付"
    and month(paidTime) = 3
    group by userid) as user_3
inner join (
    select userid from test1.orderinfo
    where isPaid = "已支付"
    and month(paidTime) = 4
    group by userid) as user_4 
on user_3.userid = user_4.userid;

因此最后： 13119/54799 =23.94%， 所以3月回购率是23.94%。

2.3计算所有月份的回购率

需要对2.2的代码进行修改。

select * from (
    select userid, date_format(paidTime, "%Y-%m-01") as m, count(isPaid) from test1.orderinfo
    where isPaid = "已支付"
    group by userid, m)) as a
left join (
    select userid, date_format(paidTime, "%Y-%m-01") as m, count(isPaid) from test1.orderinfo
    where isPaid = "已支付"
    group by userid, m) as b
on a.userid = b.userid 

上面的代码把两个完全相同的表连接，解释：

1.子查询使用userid和date_format(paidTime, "%Y-%m-01")列进行分组。即得到的记录是用户在一个月内的消费次数。

2.使用date_format()函数，改变格式，把日设为01，以便进行后面的DATE_SUB(start_date,INTERVAL expr unit)计算。

然后，见完整代码：

3.⚠️使用的是左连接left join

4.因为使用左连接，所以可以这么增加一个筛选条件：and a.m = date_sub(b.m, interval 1 month)

•  即让a的月份=b的月份-1.   这样就可以把：每条记录按照同一用户在3月消费记录和4月消费记录连接起来，同样4月和5月，6月和7月等等，即相邻月份进行关联，以便计算回购率。
•  不符合条件的b表的value都是null表示。

5. 最后的外层表用a.m进行分组，然后使用聚合函数统计a表每个月的消费人数，以及b.表每个月的消费人数。因为通过条件合并了相邻月份。所以最后得到回购率的分母和分子。

select a.m, count(a.m),count(b.m) from (
    select userid, date_format(paidTime, "%Y-%m-01") as m, count(isPaid) from test1.orderinfo
    where isPaid = "已支付"
    group by userid, m) as a
left join (
    select userid, date_format(paidTime, "%Y-%m-01") as m, count(isPaid) from test1.orderinfo
    where isPaid = "已支付"
    group by userid, m) as b
on a.userid = b.userid 
and a.m = date_sub(b.m, interval 1 month)
group by a.m

 ⚠️

6. 这里必须使用and,而不是where,

• 因为where是对left join ...on..后的数据再筛选，不符合条件的就被去掉了。
• 而and是在 on子句的内部，不符合条件的数据会用null表示。

and a.m = date_sub(b.m, interval 1 month)

---

  

3.统计男女的消费频次是否有差异

设所有男性消费者为x, 所有男性消费者的合计消费次数是x_order，那么男性消费频次为 x_order/x

select sex, avg(ct) from (
    select o.userid, t.sex, count(1) as ct from orderinfo as o
    inner join (
        SELECT * FROM test1.userinfo
        where sex <> "") t
    on o.userid = t.userid and o.ispaid = "已支付"
    group by o.userid, t.sex) t2
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

(相当于一个消费者的消费的周期)

首先，查询每个用户第一次消费和最后一次消费的时间，使用max, min函数：

SELECT userid, max(paidtime), min(paidtime) FROM test1.orderinfo
where ispaid = "已支付"
group by userid
having count(1) > 1

然后，让max()减去min()但得到的是秒，所以需要使用datediff()函数，

SELECT userid,max(paidtime), min(paidtime), datediff(max(paidtime), min(paidtime)) as interval_day FROM test1.orderinfo
where ispaid = "已支付"
group by userid having count(1) > 1

最后，再加上一层：

select avg(df) from (
    SELECT userid,max(paidtime), min(paidtime), datediff(max(paidtime), min(paidtime)) as df FROM test1.orderinfo
    where ispaid = "已支付"
    group by userid
    having count(1) > 1) t

得到一个结果：15.6484天。

求平均值，这是不准确的简单计算。根据用户类型，个别高消费频次用户，应该属于统计中的极值，不当算在统计样本中。

或者改求中位数的值。

---

  

5统计不同年龄段，用户的消费金额是否有差异？(将消费金额定义为每个年龄段的人均消费金额和总金额。)

首先，得到一个每个用户所在年龄组的表t1

然后，使用orderinfo，得到每个用户的消费总额的表t2。

最后，把t1,t2内连接后，按照age_group分组，并使用聚合函数avg()计算每个年龄组的人均消费金额。

select t1.age_group, cast(avg(t2.price) as decimal(10,2)) as avg_price from
    (select userid, ceil((year(now())-year(birth))/10)  as age_group from userinfo
    where birth > '1901-01-01') t1　　　　　　　　　　　　　　#用'1901-01-01'去除一些脏数据
inner join (
    SELECT userid, sum(price) as price FROM test1.orderinfo
    where isPaid = '已支付'
    group by userid) t2
on t1.userid = t2.userid
group by t1.age_group order by age_group

更好的方法是使用case when：

select t1.age_group, cast(avg(t2.price) as decimal(10,2)) as avg_price from (
    select userid,
    case 
        when (year(now())-year(birth)) <=10 then "<=10 "
        when (year(now())-year(birth)) between 11 and 20 then "10 to 20"
        when (year(now())-year(birth)) between 21 and 30 then "20 to 30"
        when (year(now())-year(birth)) between 31 and 40 then "30 to 40"
        when (year(now())-year(birth)) between 41 and 50 then "40 to 50"
        when (year(now())-year(birth)) between 51 and 70 then "50 to 70"
        when (year(now())-year(birth)) >=71 then ">=71"
    end as age_group
    from userinfo
    where birth > '1901-01-01') t1
inner join (
    SELECT userid, sum(price) as price FROM test1.orderinfo
    where isPaid = '已支付'
    group by userid) t2
on t1.userid = t2.userid
group by t1.age_group order by t1.age_group 

结果：

+-----------+-----------+
| age_group | avg_price |
+-----------+-----------+
| 10 to 20  |    846.63 |
| 20 to 30  |   1003.96 |
| 30 to 40  |   1178.61 |
| 40 to 50  |   1183.59 |
| 50 to 70  |   1099.86 |
| <=10      |   1322.29 |
| >=71      |   3269.92 |
+-----------+-----------+

---

6统计消费的2/8法则，消费的top20%的用户，贡献了多少额度

首先，计算出每个用户的消费额度@total, 和总共有多少消费用户。

然后，计算前20%的用户数量，使用limit得到这个用户范围的表格t

最后，对t计算消费总额@top20_price，这个数值除以@total,得到0.85。即前20%的用户，贡献了85%的额度。

select @count := count(userid), @total := sum(price) as total from (
    SELECT userid, sum(price) as price FROM test1.orderinfo
    where isPaid = '已支付'
    group by userid
    order by price desc) t; #total 318503081.54, count = 85649
 
select @top20_percent := ceil(@count*0.2); #17130 

select @top20_price :=sum(t.price) from (
    SELECT userid, sum(price) as price FROM test1.orderinfo
    where isPaid = '已支付'
    group by userid
    order by price desc
    limit 17130) t # 272203711.45
select @top20_price/@total; #0.85

或者：

select @count := count(userid), @total := sum(price) as total from (
    SELECT userid, sum(price) as price FROM test1.orderinfo
    where isPaid = '已支付'
    group by userid
    order by price desc) t; #total 318503081.54, count = 85649
--  
select @top20_percent := ceil(@count*0.2); #17130 

select sum(price)/@total from (
    select 
    row_number() over(order by price desc ) as rk,
    userid, price
    from (
        SELECT userid, sum(price) as price FROM test1.orderinfo
        where isPaid = '已支付'
        group by userid
        order by price desc) as t1
    ) as t2
where t2.rk < @top20_percent

这里使用row_numbe()计算函数，得到一个排名，然后用where子句得到前20%的用户的表t2，最后t2进行计算。