使用静态最大日期和最小日期查找每日会话的平均值答案

【问题标题】：Finding Average of Daily Sessions with a Static Max Date and Min Date使用静态最大日期和最小日期查找每日会话的平均值
【发布时间】：2017-03-22 16:50:16
【问题描述】：

PHP 5.X、MySQL 5.7.15

获取按星期几分组的数据，以使用 Max Date 和 Min Date 提供每日使用的平均会话数。在这种情况下，最大和最小日期被应用于星期几，这导致每个案例的周数不同。

生成这些结果的查询定义为下面的查询 1。本质上，最后 4 列中的信息应该是静态的，我使用 SELF JOIN 来完成此操作，详情如下。在该示例中，session_count 爆炸式增长，因为我只在开发环境中工作。

那么出了什么问题，为什么？

  session_count | weekday | maxDate             |minDate              | numOfDayDiff| numOfWeeks
'29'            | 'Friday'|'2017-03-10 12:16:47'|'2016-08-12 12:31:28'| '210'       | '30'
'26'            |'Monday' |'2017-03-06 17:10:59'|'2016-08-08 14:31:16'| '210'       | '30'
'6'            |'Saturday'|'2017-03-04 23:26:12'|'2016-08-20 23:10:47'| '196'       | '28'
'10'            |'Sunday' |'2017-03-12 18:28:51'|'2016-08-14 16:26:30'| '210'       | '30'
'22'           |'Thursday'|'2017-03-09 15:45:27'|'2016-06-30 12:42:19'| '252'       | '36'
'22'            |'Tuesday'|'2017-03-21 15:00:08'|'2016-06-28 19:51:21'| '266'       | '38'
'22'          |'Wednesday'|'2017-03-01 23:57:34'|'2016-07-06 16:17:39'| '238'       | '34'

我希望最大日期和最小日期是静态的。我试过用w3schools解释Self JOIN Syntax

SELECT column_name(s)
FROM table1 T1, table1 T2
WHERE condition;

我尝试应用此查询，maxDate、minDate、numDaysDiff 和 numOfWeeks 与上述最高周数的值一致，但 session_count 的统计数据呈爆炸式增长。

这里是Query 1，它获得了上表：

SELECT
CEIL(COUNT(ss.session_id) / (FLOOR (DATEDIFF(MAX(ss.date),MIN(ss.date))/7))) as session_count,
DAYNAME((ss.date)) as weekday,
MAX(ss.date) as maxDate,
MIN(ss.date) as minDate,
DATEDIFF(MAX(ss.date),MIN(ss.date)) as numOfDayDiff,
FLOOR(DATEDIFF(MAX(ss.date),MIN(ss.date))/7) as numOfWeeks
FROM  session ss
JOIN user u ON ss.user = u.id
JOIN account a on u.account = a.id
WHERE a.isTest=false
GROUP BY weekday
ORDER BY weekday;

这里是 Q-2 使用 SELF JOIN 语法修改后的输出表：

SELECT
CEIL(COUNT(ss.session_id) / (FLOOR (DATEDIFF(MAX(xx.date),MIN(xx.date))/7))) as session_count,
DAYNAME((ss.date)) as weekday,
MAX(xx.date) as maxDate,
MIN(xx.date) as minDate,
DATEDIFF(MAX(xx.date),MIN(xx.date)) as numOfDayDiff,
FLOOR(DATEDIFF(MAX(xx.date),MIN(xx.date))/7) as numOfWeeks
FROM  session xx, session ss
JOIN user u ON ss.user = u.id
JOIN account a on u.account = a.id
WHERE a.isTest=false
GROUP BY weekday
ORDER BY weekday;

结果是：

session_count | weekday | maxDate             |minDate              | numOfDayDiff| numOfWeeks
99357         |Friday   |2017-03-21 15:00:08  |2016-06-28 19:51:21  |266          |38
88062         |Monday   |2017-03-21 15:00:08  |2016-06-28 19:51:21  |266          |38
16829         |Saturday |2017-03-21 15:00:08  |2016-06-28 19:51:21  |266          |38
32505         |Sunday   |2017-03-21 15:00:08  |2016-06-28 19:51:21  |266          |38
90136         |Thursday |2017-03-21 15:00:08  |2016-06-28 19:51:21  |266          |38
94516         |Tuesday  |2017-03-21 15:00:08  |2016-06-28 19:51:21  |266          |38
83451         |Wednesday|2017-03-21 15:00:08  |2016-06-28 19:51:21  |266          |38

谁能告诉我我是怎么出错的？我将看看是否可以创建一个 mysql fiddle 以作为示例附加在这里。

【问题讨论】：

标签： php mysql self-join

【解决方案1】：

为了减轻数据库的压力，我决定使用 Query 1 的修改版本，并使用结果来确定哪一个的周数最多，然后将计数除以周数在php中。

这是修改后的查询1

SELECT
COUNT(ss.session_id) as session_count,
DAYNAME((ss.date)) as weekday,
MAX(ss.date) as maxDate,
MIN(ss.date) as minDate,
CEIL(DATEDIFF(MAX(ss.date),MIN(ss.date))/7) as numOfWeeks
FROM  session ss
JOIN user u ON ss.user = u.id
JOIN account a on u.account = a.id
WHERE a.isTest=false
GROUP BY weekday
ORDER BY weekday;

session_count | weekday | maxDate            | minDate           | numOfWeeks
2017          | Friday  |2017-03-17 23:39:56 |2016-09-09 01:06:36|  27
1908          | Monday  |2017-03-20 23:59:53 |2016-09-12 14:52:29|  27
943           | Saturday|2017-03-18 23:39:14 |2016-09-17 14:49:56|  26
825           | Sunday  |2017-03-19 23:56:37 |2016-09-18 22:28:45|  26
2429          | Thursday|2017-03-16 23:51:23 |2016-09-08 12:02:54|  27
2341          | Tuesday |2017-03-21 23:52:20 |2016-09-06 12:06:39|  28
2569          |Wednesday|2017-03-22 17:03:16 |2016-09-07 02:43:10|  28

注意数据因数据库不同而不同。

【讨论】：