231 lines
5.7 KiB
Markdown
231 lines
5.7 KiB
Markdown
# SQL窗口函数（二）—— 连续问题
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### 题目一
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#### 表格
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| user_id | login_date |
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| ------- | ---------- |
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| A | 2022-09-02 |
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| A | 2022-09-03 |
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| A | 2022-09-04 |
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| B | 2021-11-25 |
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| B | 2021-12-31 |
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| C | 2022-01-01 |
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| C | 2022-04-04 |
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| C | 2022-09-03 |
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| C | 2022-09-04 |
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| C | 2022-09-05 |
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| A | 2022-09-03 |
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| D | 2022-10-20 |
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| D | 2022-10-21 |
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| A | 2022-10-03 |
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| D | 2022-10-22 |
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| D | 2022-10-23 |
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#### 脚本
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```sql
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CREATE TABLE SQL_8
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(
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user_id varchar(2),
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login_date date
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);
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INSERT INTO SQL_8 (user_id,login_date)
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VALUES ('A', '2022-09-02'), ('A', '2022-09-03'), ('A', '2022-09-04'), ('B', '2021-11-25'),
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('B', '2021-12-31'), ('C', '2022-01-01'), ('C', '2022-04-04'), ('C', '2022-09-03'),
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('C', '2022-09-05'), ('C', '2022-09-04'), ('A', '2022-09-03'), ('D', '2022-10-20'),
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('D', '2022-10-21'), ('A', '2022-10-03'), ('D', '2022-10-22'), ('D', '2022-10-23');
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```
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#### 问题
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找出这张表中所有的连续3天登录用户
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#### 分析
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连续N天登录用户，要求数据行满足以下条件：
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1) userid 要相同，表示同一用户
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2) 同一用户每行记录以登录时间从小到大排序
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3) 后一行记录比前一行记录的登录时间多一天
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4) 数据行数大于等于N
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#### 解答
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```sql
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-- 方法一
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with t1 as (
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select distinct user_id, login_date from SQL_8
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),
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t2 as (
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select *, row_number() over (partition by user_id order by login_date) as rn from t1
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),
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t3 as (
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select *, DATE_SUB(login_date, interval rn day) as sub from t2
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)
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select distinct user_id from t3 group by user_id, sub having count(user_id) >= 3;
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-- 方法二
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with t1 as (
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select distinct user_id, login_date from SQL_8
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),
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t2 as (
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select *, DATEDIFF(login_date, lag(login_date, 1) over (partition by user_id order by login_date)) as diff from t1
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)
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select user_id from t2 where diff = 1 group by user_id having count(user_id) >= 2;
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```
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### 题目二
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#### 表格
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| player_id | score | score_time |
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| --------- | ----- | ------------------- |
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| B3 | 1 | 2022-09-20 19:00:14 |
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| A2 | 1 | 2022-09-20 19:01:04 |
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| A2 | 3 | 2022-09-20 19:01:16 |
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| A2 | 3 | 2022-09-20 19:02:05 |
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| A2 | 2 | 2022-09-20 19:02:25 |
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| B5 | 2 | 2022-09-20 19:02:54 |
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| A4 | 3 | 2022-09-20 19:03:10 |
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| B1 | 2 | 2022-09-20 19:03:34 |
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| B1 | 2 | 2022-09-20 19:03:58 |
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| B1 | 3 | 2022-09-20 19:04:07 |
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| A2 | 1 | 2022-09-20 19:04:19 |
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| A3 | 2 | 2022-09-20 19:04:31 |
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#### 脚本
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```sql
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CREATE TABLE SQL_9
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(
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player_id varchar(2),
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score int,
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score_time datetime
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);
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INSERT INTO SQL_9 (player_id, score, score_time)
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VALUES ('B3', 1, '2022-09-20 19:00:14'), ('A2', 1, '2022-09-20 19:01:04'),
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('A2', 3, '2022-09-20 19:01:16'), ('A2', 3, '2022-09-20 19:02:05'),
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('A2', 2, '2022-09-20 19:02:25'), ('B3', 2, '2022-09-20 19:02:54'),
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('A4', 3, '2022-09-20 19:03:10'), ('B1', 2, '2022-09-20 19:03:34'),
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('B1', 2, '2022-09-20 19:03:58'), ('B1', 3, '2022-09-20 19:04:07'),
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('A2', 1, '2022-09-20 19:04:19'), ('B3', 2, '2022-09-20 19:04:31');
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```
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#### 问题
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统计出连续三次（及以上）为球队得分的球员名单
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#### 分析
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连续N次以上为球队得分， 要求数据行满足以下条件：
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1) player_id 要相同表示同一球员
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2) 每行记录以得分时间从小到大排序
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3) 数据行数大于等于N
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#### 解答
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```sql
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-- 方法一
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with t1 as (
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select *, lag(player_id, 1) over (order by score_time) as last_play_id from SQL_9
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)
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select distinct player_id from t1 where player_id = last_play_id group by player_id having count(player_id) >= 2;
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-- 方法二
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with t1 as (
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select *, row_number() over (order by score_time) as rn from SQL_9
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),
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t2 as (
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select *, row_number() over (order by score_time) + 1 as rn from SQL_9
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),
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t3 as (
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select t1.player_id as player_id from t1 join t2 on t1.rn = t2.rn and t1.player_id = t2.player_id
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)
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select distinct player_id from t3 group by player_id having count(player_id) >= 2;
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```
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### 题目三
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#### 表格
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| log_id |
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| :----: |
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| 1 |
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| 2 |
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| 3 |
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| 7 |
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| 8 |
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| 10 |
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#### 脚本
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```sql
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CREATE TABLE SQL_10
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(
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log_id int
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);
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INSERT INTO SQL_10 (log_id) VALUES (1), (2), (3), (7), (8), (10);
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```
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#### 问题
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编写SQL 查询得到 Logs 表中的连续区间的开始数字和结束数字。按照 start_id 排序。查询结果格式如下:
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| start_id | end_id |
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| -------- | ------ |
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| 1 | 3 |
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| 7 | 8 |
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| 10 | 10 |
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#### 解答
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```sql
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-- 方法一
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with t1 as (
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select *, log_id - row_number() over (order by log_id) as gr from SQL_10
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)
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select min(log_id) as start_id, max(log_id) as end_id from t1 group by gr
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-- 方法二
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with t1 as (
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select *, log_id - row_number() over (order by log_id) as gr, log_id - lag(log_id,1) over () as diff from SQL_10
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),
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t2 as (
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select log_id, gr from t1 where ifnull(diff,-1) <> 1
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),
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t3 as (
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select *, log_id - row_number() over (order by log_id) as gr, log_id - lead(log_id,1) over () as diff from SQL_10
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),
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t4 as (
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select log_id, gr from t3 where ifnull(diff, 1) <> -1
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)
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select t2.log_id as start_id, t4.log_id as end_id from t2, t4 where t2.gr = t4.gr;
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```
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### 技巧
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如何求连续区间？
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1）行号过滤法
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通过row_number() 生成连续行号，与区间列进行差值运算，得到的临时结果如果相同表示为同一连续区间
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2） 错位比较法
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通过row_number() / row_number() + 1 分别生成原生的和错位的连续行号列，进行连表操作
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也可以通过lag/lead函数直接生成错位列 |