SQL：如何计算没有GROUP BY的值的总和

Question

我有下表：

visitorId    visitNumber   DATE         
1            1             20180101     
1            2             20180101   
1            3             20180105    
2            1             20171230  
2            2             20180106    

我想返回的是：

visitorId    totalVisits   max_visits_in_1_day  
1            3             2                       
2            2             1                       

我设法使用max_visits_in_1_day来使一切正常max_visits_in_1_day ：

SELECT visitorId, 
MAX(visitNumber) - MIN(visitNumber) + 1 as totalVisits, 
GROUP BY visitorId

我需要做的是改进代码， max_visits_in_1_day添加max_visits_in_1_day 。 像MAX(COUNT(GROUP BY(DATE)))

我首先尝试添加MAX(COUNT(DATE)) ，但这会汇总所有日期，并且实际上并没有寻找最大的唯一日期。 从某种意义上说，我需要在DATE进行GROUP BY，然后将计数加起来。

我尝试添加GROUP BY visitorId, DATE但这会创建额外的行。

Answer 1

您将必须执行以下两个步骤：

SELECT visitorId, SUM(perDay) AS totalVisits, MAX(perDay) AS max_visits_in_1_day
FROM
(SELECT visitorId, COUNT(visitNumber) AS perDay, DATE
FROM myTable
GROUP BY visitorId, DATE) A
GROUP BY visitorId

Answer 2

您可以尝试以下查询-

SELECT visitorId
      ,COUNT(visitNumber) totalVisits
      ,mv1d.count max_visits_in_1_day
FROM YOUR_TABLE YT
INNER JOIN (SELECT visitorId, MAX(COUNT(DATE)) count
            FROM YOUR_TABLE YT1)
ON YT.visitorId = YT1.visitorId
GROUP BY visitorId