mysql - 基于mysql中的周数求平均值

Question

我正在尝试完成一个查询，该查询需要两个指标的平均值:入站调用和未接调用。但我从来没有如此细致地处理过星期几和每小时的数据，所以我不确定我的总计是否正确，更不用说得到正确的平均值了。

基本上，从 2018 年 1 月 1 日开始，我想要周一至周五上午 7 点到下午 6 点每小时的平均来电和未接来电。我们关闭了 7 天，我得到了 48 行，所以这就是我的预期。

如果过去 3 个周一是这样的:

creationtimestamp   | Legtype1  |  answered  

07/23/18 08:15:00   |     2     |     0             
07/23/18 08:25:00   |     2     |     1           
07/23/18 08:35:00   |     2     |     1               
07/30/18 08:15:00   |     2     |     0           
07/30/18 08:25:00   |     2     |     0           
07/30/18 08:35:00   |     2     |     0           
07/30/18 08:45:00   |     2     |     1           
07/30/18 08:55:00   |     2     |     0           
08/06/18 08:15:00   |     2     |     0           
08/06/18 08:25:00   |     2     |     1           
08/06/18 08:35:00   |     2     |     0           
08/06/18 08:45:00   |     2     |     0  

周一上午 8 点到 9 点期间总共接到 12 个电话，其中 4 个未接。如果我从 8 点到 9 点查询这三个星期一，我会期望:

Monday | 8 | 4 | 1.3

但是我不知道如何计算每个工作日的所有调用的总和，将它们相加并除以该工作日的数量？我下面的查询目前有“SUM”而不是平均值，但我不确定如何获取我需要的平均值，因为它取决于每个工作日的数字。

SELECT 
    dayname(s.creationtimestamp) as day, -- weekdays
    HOUR(s.creationtimestamp) as Hour, -- Hours
    sum(case when legtype1 = 2 then 1 else 0 end) as total_calls, -- total inbound
    sum(case when  legtype1 = 2 and answered = 0 then 1 else 0 end)as total_missed

FROM session s
WHERE (s.creationtimestamp >= '2018-01-01' AND creationtimestamp < now())
and WEEKDAY(s.creationtimestamp) BETWEEN 0 AND 4 -- Monday through friday
AND HOUR(s.creationtimestamp) between  7 and 18 -- 7am to 6pm
GROUP BY dayname(s.creationtimestamp),  HOUR(s.creationtimestamp)
order by dayofweek(s.creationtimestamp), hour asc;

重申一下:查询有效，但我不确定是否根据从今年第一天到现在的每个工作日和小时 block 正确聚合。

这是一个 fiddle : http://sqlfiddle.com/#!9/7b6b72

Answer 1

我认为一切对你来说都很好，只是多一点，希望下面的查询对你有帮助

select day,Hour,Avg(total_calls) as avg_total_calls,
Avg(total_missed) as avg_total_missed from
(
SELECT 
    dayname(s.creationtimestamp) as day, -- weekdays
    HOUR(s.creationtimestamp) as Hour, -- Hours
    sum(case when legtype1 = 2 then 1 else 0 end) as total_calls, -- total inbound
    sum(case when  legtype1 = 2 and answered = 0 then 1 else 0 end)as total_missed

FROM session s
WHERE (s.creationtimestamp >= '2018-01-01' AND creationtimestamp < now())
and WEEKDAY(s.creationtimestamp) BETWEEN 0 AND 4 -- Monday through friday
AND HOUR(s.creationtimestamp) between  7 and 18 -- 7am to 6pm
GROUP BY dayname(s.creationtimestamp),  HOUR(s.creationtimestamp)

) as T group by day,Hour 
order by day,Hour asc