MySQL连续数天/连续天数

Question

我有两张桌子。两者都包含一个用户的条目，一个来自早上，一个来自晚上。

我试图获取天数，因此用户在早上和晚上添加了一个条目。每当有一个缺失时，“连胜”就会结束——在当天结束。所以它不介意用户是否在一年前活跃并且连续100天，它只是“用户连续多少天在morningTable和eveningTable中添加条目，返回从今天开始”。

两个表都有每个条目的时间戳。

一个例子：

morningTable:

ID text timestamp
1  bla  2020-08-07 10:30:00
2  blub 2020-08-06 11:30:00
3  abc  2020-08-05 09:20:00
4  def  2020-08-04 06:13:00
5  ghi  2020-08-02 07:05:00

eveningTable:

ID text timestamp
1  abc  2020-08-07 19:30:00
2  def  2020-08-06 20:45:00
3  ghi  2020-08-05 21:30:00
4  jkl  2020-08-03 20:13:00
5  mno  2020-08-01 17:33:00
6  pqr  2020-07-29 19:19:19

因此，对于这两个表，结果必须是数字 3，因为两个表都连续 05.08 - 07.08...

希望你能理解我的情况。

我找到了带有 ROWNUMBER 的示例来查找差距等，但是，很难将它与两个表结合起来。

谁能帮我看看如何用 MySQL 做到这一点？

虽然我也只是在 PHP 中收集所有数据并在那里执行，但是，当数据增长时，使用 MySQL（数据库端）执行它会比在 PHP 中执行要好得多。

提前致谢！

Answer 1

使用union all 组合表格并聚合以得到天数both。然后使用 gaps-and-islands 方法：

with me as (
      select dte
      from ((select date(timestamp) as dte, 1 as morning, 0 as evening
             from morning
            ) union all
            (select date(timestamp) as dte, 0 as morning, 1 as evening
             from evening
            )
           ) me
      group by dte
      having sum(morning) > 0 and sum(evening) > 0
     )
select min(dte), max(dte), count(*)
from (select dte, row_number() over (order by dte) as seqnum
      from me
     ) me
group by dte - interval seqnum day
order by count(*) desc
limit 1;

Here 是一个 dbfiddle。

编辑：

其实这不是特别难为这个问题调整变量：

select min(dte), max(dte), count(*)
from (select dte, (@rn := @rn + 1) as seqnum
      from (select dte
            from ((select date(timestamp) as dte, 1 as morning, 0 as evening
                   from morning
                  ) union all
                  (select date(timestamp) as dte, 0 as morning, 1 as evening
                   from evening
                  )
                 ) me
            group by dte
            having sum(morning) > 0 and sum(evening) > 0
            order by dte
           ) d cross join
           (select @rn := 0) params
     ) me
group by dte - interval seqnum day
order by count(*) desc
limit 1;

Answer 2

WITH 
cte1 AS ( SELECT DISTINCT DATE(m.`timestamp`) uni_date                              
          FROM morningTable m
          JOIN eveningTable e ON DATE(m.`timestamp`) = DATE(e.`timestamp`) ),
cte2 AS ( SELECT uni_date, ROW_NUMBER() OVER (ORDER BY uni_date DESC) rn
          FROM cte1 )
SELECT MAX(DATEDIFF( CURRENT_DATE, uni_date ) + 1) output
FROM cte2
WHERE rn = DATEDIFF( CURRENT_DATE, uni_date ) + 1

如果表格包含太多行，那么使用某个日期值限制在cte1 中检查的日期可能是安全的，该日期值在后一个保证间隔之前。

PS。抱歉，这从 MariaDB 10.2.1 开始可用...

---

在 MariaDB 10.1.45 中，下一个查询必须有效：

SELECT DATEDIFF(CURRENT_DATE, MIN(uni_date)) + 1 output
FROM ( SELECT DISTINCT DATE(m.`timestamp`) uni_date                              
       FROM morningTable m
       JOIN eveningTable e ON DATE(m.`timestamp`) = DATE(e.`timestamp`) ) t1
WHERE NOT EXISTS ( SELECT NULL
                   FROM ( SELECT DISTINCT DATE(m.`timestamp`) uni_date                              
                          FROM morningTable m
                          JOIN eveningTable e ON DATE(m.`timestamp`) = DATE(e.`timestamp`) ) t2
                   WHERE t2.uni_date = t1.uni_date - INTERVAL 1 DAY )
  AND ( SELECT DATE(MAX(m.`timestamp`))
        FROM morningTable m
        JOIN eveningTable e ON DATE(m.`timestamp`) = DATE(e.`timestamp`) ) = CURRENT_DATE

如果今天在至少一个表中没有行，则查询将返回 NULL - 您可以使用 COALESCE for 来获取零。