如何获得总计数值每天最多5天

cygmwpex  于 2021-06-21  发布在  Mysql
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"SELECT count(id) AS total FROM participant where dateofbooking='$datepick'";

我正在使用这个代码。但它只显示一个日期(从php中选择的日期)计数。但我想选择单一的日期,它应该显示我多达5天每天明智的计数预订。
输出应该是this:-

2018-05-20------>48
2018-05-21------>58
2018-05-22------>67
2018-05-23------>78
2018-05-24------>43
toe95027

toe950271#

您可以按使用的日期列对其进行分组,如果需要多天,可以添加 dateofbooking >= some_start_date 以及 dateofbooking <= some_end_date ```
"SELECT count(id) AS total FROM participant where dateofbooking='$datepick' group by dateofbooking";

倍数可能看起来像

"SELECT count(id) AS total FROM participant where dateofbooking>='$datepickstart' AND dateofbooking<='$datepickend' group by dateofbooking";

qhhrdooz

qhhrdooz2#

你可以用 between . 请参见下面的示例:

SELECT count(id) AS total FROM participant where dateofbooking between cast(  GETDATE() as Date) and Cast(DATEADD(DAY,-5, GETDATE()) as DATE)
Group by CAST( dateofbooking AS DATE)

另外,使用 group by Date .
注意:这不是mysql的解决方案。它适用于sql server。

cl25kdpy

cl25kdpy3#

你可以用 DATE_ADD() :

SELECT dateofbooking, count(id) AS total 
FROM participant 
WHERE dateofbooking >= $datepick AND 
      dateofbooking <= DATE_ADD($datepick, INTERVAL 5 DAY)
GROUP BY dateofbooking;

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