SQL实战：05之间隔连续数问题求解
概述
最近刷题时遇到一些比较有意思的题目，之前多次遇到一些求解连续数的问题，这次遇到了他们的变种，连续数可以间隔指定的数也视为是一个完整的“连续”。针对连续数的这类问题我们之前讲的可以利用等差数列的思想来解决，然而现在发生了变形一定还能继续使用等差数列来解决吗？答案是否定的，针对带有指定间隔的连续数问题应当使用分组问题的思想来解决。
题目：间隔连续表
某游戏公司记录的用户每日登录数据
表logs
| 字段名 | 数据类型 |
|---|---|
| id | bigint |
| dt | date |
输入数据如下：
| id | dt |
|---|---|
| 1001 | 2021/12/12 |
| 1002 | 2021/12/12 |
| 1001 | 2021/12/13 |
| 1001 | 2021/12/14 |
| 1001 | 2021/12/16 |
| 1002 | 2021/12/16 |
| 1001 | 2021/12/19 |
| 1002 | 2021/12/17 |
| 1001 | 2021/12/20 |
计算每个用户最大的连续登录天数,可以间隔一天。解释:如果一个用户在 1,3,5,6 登
录游戏,则视为连续 6 天登录。
题解
第一步：下移数据
WITH temp_001 AS (SELECT id,dt,LAG(dt,1,'1970-01-01') OVER (PARTITION BY id ORDER BY dt ASC) AS prev_dtFROM logs
)
输出如下：
| id | dt | prev_dt |
|---|---|---|
| 1001 | 2021/12/12 | 1970/01/01 |
| 1001 | 2021/12/13 | 2021/12/12 |
| 1001 | 2021/12/14 | 2021/12/13 |
| 1001 | 2021/12/16 | 2021/12/14 |
| 1001 | 2021/12/19 | 2021/12/19 |
| 1001 | 2021/12/20 | 2021/12/20 |
| 1002 | 2021/12/12 | 1970/01/01 |
| 1002 | 2021/12/16 | 2021/12/12 |
| 1002 | 2021/12/17 | 2021/12/16 |
步骤二：窗口分组累加
计算当前日期和上一日期的差值，如果差值小于等于2，则认为是同一组组号不变，如果差值大于2则任务是新的组，组号+1.
SQL实现：
temp_002 AS (SELECT id,dt,SUM(IF(DATEDIFF(dt,prev_dt)>2,1,0)) OVER(PARTITION BY id ORDER BY dt ASC) AS group_idFROM temp_001
)输出结果：
| id | dt | group_id |
|---|---|---|
| 1001 | 2021/12/12 | 1 |
| 1001 | 2021/12/13 | 1 |
| 1001 | 2021/12/14 | 1 |
| 1001 | 2021/12/16 | 1 |
| 1001 | 2021/12/19 | 2 |
| 1001 | 2021/12/20 | 2 |
| 1002 | 2021/12/12 | 1 |
| 1002 | 2021/12/16 | 2 |
| 1002 | 2021/12/17 | 2 |
步骤三：根据id和group分组统计出最大、最小日期
temp_003 AS (SELECT id,group_id,min(dt) as min_dt,max(dt) as max_dtFROM temp_002GROUP BY id,group_id
)
输出结果
| id | group_id | min_dt | max_dt |
|---|---|---|---|
| 1001 | 1 | 2021/12/12 | 2021/12/16 |
| 1001 | 2 | 2021/12/19 | 2021/12/20 |
| 1002 | 1 | 2021/12/12 | 2021/12/12 |
| 1002 | 2 | 2021/12/16 | 2021/12/17 |
步骤四：使用最大日期减掉最小日期再加1，就可以得到每一组的最大连续天数
temp_004 AS (SELECT id,group_id,min_dt,max_dt,datediff(max_dt,min_dt) + 1 AS daysFROM temp_003
)
输出结果
| id | group_id | min_dt | max_dt | days |
|---|---|---|---|---|
| 1001 | 1 | 2021/12/12 | 2021/12/16 | 5 |
| 1001 | 2 | 2021/12/19 | 2021/12/20 | 2 |
| 1002 | 1 | 2021/12/12 | 2021/12/12 | 1 |
| 1002 | 2 | 2021/12/16 | 2021/12/17 | 2 |
步骤五： 在取每个id的最大连续天数得到最终结果
SELECT id,max(days) as days from temp_004 group by id;
完整SQL
WITH temp_001 AS (SELECT id,dt,LAG(dt,1,'1970-01-01') OVER (PARTITION BY id ORDER BY dt ASC) AS prev_dtFROM logs
)
,temp_002 AS (SELECT id,dt,SUM(IF(DATEDIFF(dt,prev_dt)>2,1,0)) OVER(PARTITION BY id ORDER BY dt ASC) AS group_idFROM temp_001
)
,temp_003 AS (SELECT id,group_id,min(dt) as min_dt,max(dt) as max_dtFROM temp_002GROUP BY id,group_id
)
,temp_004 AS (SELECT id,group_id,min_dt,max_dt,datediff(max_dt,min_dt) + 1 AS daysFROM temp_003
)
SELECT id,max(days) as days from temp_004 group by id;