使用组合表格 union all 加起来，两个都有好几天了。然后使用间隙和孤岛方法：

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;

这是一把小提琴。
编辑：
实际上，要使变量适应这个问题并不特别困难：

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;

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

如果表包含太多的行，那么限制签入的日期可能是安全的 cte1 有一些日期值，在后一个与保证的差距之前。
抱歉，这是从10.2.1开始提供的。。。
在10.1.45中，下一个查询必须工作：

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

如果今天至少有一个表中没有行，那么查询将返回null—您可以使用coalesce for来获得零。