SQL的两道练习

发表于 2017-07-05 更新于 2025-01-16 分类于数据库 Waline：阅读次数：

找了两道Hive有关的SQL练习题目，第一道是统计用户浏览量的数据。第二道是统计京东商铺浏览量的数据。

（一）统计用户浏览量

我们有如下的用户访问数据

userId	visitDate	visitCount
u01	2017/1/21	5
u02	2017/1/23	6
u03	2017/1/22	8
u04	2017/1/20	3
u01	2017/1/23	6
u01	2017/2/21	8
u02	2017/1/23	6
u01	2017/2/22	4

要求使用SQL统计出每个用户的累积访问次数，如下表所示：

用户id	月份	小计	累积
u01	2017-01	11	11
u01	2017-02	12	23
u02	2017-01	12	12
u03	2017-01	8	8
u04	2017-01	3	3

原始数据如下：

u01     2017/1/21       5
u02     2017/1/23       6
u03     2017/1/22       8
u04     2017/1/20       3
u01     2017/1/23       6
u01     2017/2/21       8
u02     2017/1/23       6
u01     2017/2/22       4

建表语句

create table action
(userId string,
visitDate string,
visitCount int) 
row format delimited fields terminated by "\t";

答：

一、首先转换月份格式，统计每个月、每个用户的访问量

SELECT 
    userid id , 
    date_format(REGEXP_REPLACE(visitdate,'/','-'),'yyyy-MM') mon , 
    sum(visitcount) views
FROM 
    action
GROUP BY 
    userid , visitdate;

执行结果:

id	mon	views
u01	2017-01	5
u01	2017-01	6
u01	2017-02	8
u01	2017-02	4
u02	2017-01	12
u03	2017-01	8
u04	2017-01	3

二、再按照月份和用户进行分组并统计访问量。

SELECT 
    id , 
    mon , 
    sum(visitcount) views
FROM(SELECT 
    userid id , 
    date_format(REGEXP_REPLACE(visitdate,'/','-'),'yyyy-MM') mon , 
    visitcount
FROM 
    action) t1
GROUP BY id ,mon;

执行结果

id	mon	views
u01	2017-01	11
u01	2017-02	12
u02	2017-01	12
u03	2017-01	8
u04	2017-01	3

三、对每个用户的数据进行累加，再次子查询

select
    userid,
    mon,
    views,
    (sum(views) over(partition by userid order by mon)) total
FROM
    (SELECT 
        userid , 
        mon , 
        sum(visitcount) views
    FROM(SELECT 
            userid , 
            date_format(REGEXP_REPLACE(visitdate,'/','-'),'yyyy-MM') mon , 
            visitcount
        FROM 
            action) t1
        GROUP BY 
            userid ,mon)t2;

执行结果

userid	mon	views	total
u01	2017-01	11	11
u01	2017-02	12	23
u02	2017-01	12	12
u03	2017-01	8	8
u04	2017-01	3	3

（二）统计网店访问量

有50W个京东店铺，每个顾客访问任何一个店铺的任何一个商品时都会产生一条访问日志，访问日志存储的表名为visit，访客的用户id为user_id，被访问的店铺名称为shop，请统计：
原始数据如下：

u1	a
u2	b
u1	b
u1	a
u3	c
u4	b
u1	a
u2	c
u5	b
u4	b
u6	c
u2	c
u1	b
u2	a
u2	a
u3	a
u5	a
u5	a
u5	a

建表：

1 2	create table visit(user_id string,shop string) row format delimited fields terminated by '\t';

问：
1）每个店铺的UV（访客数）
2）每个店铺访问次数top3的访客信息。输出店铺名称、访客id、访问次数

答：

1）

一、使用GROUP BY对用户进行去重

SELECT
        shop,
        user_id
    FROM
        visit
    GROUP BY 
        shop,
        user_id

执行结果：

shop	user_id
a	u1
a	u2
a	u3
a	u5
b	u1
b	u2
b	u4
b	u5
c	u2
c	u3
c	u6

二、对去重的结果进行用户数目统计

SELECT 
    shop,
    COUNT(*) uv
FROM
    (SELECT
        shop,
        user_id
    FROM
        visit
    GROUP BY 
        shop,
        user_id) t1
GROUP BY
    shop;

执行结果

shop	uv
a	4
b	4
c	3

2）

一、计算每个人访问每个店铺的总次数

SELECT 
    shop,
    user_id,
    count(*)
FROM
    visit
GROUP BY
    shop,
    user_id;

二、针对同一店铺，对方问次数进行逆序排序，并计算rank值

SELECT
    shop,
    user_id,
    ct,
    row_number() over(partition by shop order by ct desc ) rk
FROM
    (SELECT 
        shop,
        user_id,
        count(*) ct
    FROM
        visit
    GROUP BY
        shop,
        user_id) t1;

三、去店铺访问前三名的用户

SELECT
    shop,
    user_id,
    ct
FROM
    (SELECT
        shop,
        user_id,
        ct,
        row_number() over(partition by shop order by ct desc ) rk
    FROM
        (SELECT 
            shop,
            user_id,
            count(*) ct
        FROM
            visit
        GROUP BY
            shop,
            user_id) t1)t2
WHERE
    rk <= 3;

3

Exer4：

CREATE TABLE user_age
(dt STRING, //日期
user_id STRING, //用户
age INT) //年龄

1、所有用户的总数及平均年龄

SELECT 
    user_id,
    AVG(age) age
FROM 
    user_age
GROUP BY 
    user_id;t1


SELECT
    COUNT(*) count1,
    AVG(aage) avg_total_age
FROM
    (SELECT 
    user_id,
    AVG(age) aage
FROM 
    user_age
GROUP BY 
    user_id)t1;t0

count1  avg_total_age
3   27.0
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2、活跃用户的总数及其平均年龄(2种方法)
① 用lead或者tag,在用Datediff找相邻一天的，然后去重就得出结果，不具备扩展性，实现多天活跃比较复杂。
② 等差数列
2.1 给每个用户的日期,同一天去重rank
```sql
SELECT
    user_id,
    dt,
    ROW_Number() OVER(PARTITION BY user_id ORDER BY dt) rk
FROM 
    user_age
GROUP BY
    user_id,dt;t2
```


2.2 用等差数列
```sql
SELECT
    user_id,
    DATE_SUB(dt,rk) diff
FROM
    t2;t3
```

2.3 求出差距
```sql
SELECT
    user_id,
    COUNT(diff) OVER(PARTITION BY user_id,diff) count_diff
FROM
    t3;t4
```

2.4 最大连续天数
```sql
SELECT
    user_id
FROM
    t4
WHERE
    count_diff >= 2
GROUP BY
    user_id;t5
```

2.5 去重并且得到age
```sql
SELECT
    user_age.user_id,
    user_age.age
FROM
    user_age,t5
WHERE
    user_age.user_id = t5.user_id;t6

SELECT
    COUNT(*) active_count,
    AVG(age) active_avg_age
FROM
    t6; t7
```

3、合并

最终SQL：
```sql
SELECT
    count_total,
    avg_total_age,
    active_count,
    active_avg_age
FROM
(SELECT
    COUNT(*) count_total,
    AVG(aage) avg_total_age
FROM
    (SELECT 
    user_id,
    AVG(age) aage
FROM 
    user_age
GROUP BY 
    user_id)t1)t0
,
(SELECT
    COUNT(*) active_count,
    AVG(age) active_avg_age
FROM
(SELECT
    DISTINCT user_age.user_id,
    user_age.age
FROM
    user_age,
    (SELECT
    user_id
FROM
    (SELECT
    user_id,
    COUNT(diff) OVER(PARTITION BY user_id,diff) count_diff
FROM
    (SELECT
    user_id,
    DATE_SUB(dt,rk) diff
FROM
    (SELECT
    user_id,
    dt,
    ROW_Number() OVER(PARTITION BY user_id ORDER BY dt) rk
FROM 
    user_age
GROUP BY
    user_id,dt)t2)t3)t4
WHERE
    count_diff >= 2
GROUP BY
    user_id)t5
WHERE
    user_age.user_id = t5.user_id)t6)t7;
```


~~~~~~~~~~~结果~~~~~~~~~~~
count_total avg_total_age   active_count    active_avg_age
3   27.0    1   19.0