Question

"SELECT count(id) AS total FROM participant where dateofbooking='$datepick'";

我正在使用此代码。但是它只显示一个日期（从php中选择日期）计数。但是我想选择一个日期，它应该可以显示每天最多5天的明智预订。

输出应如下所示：-

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

Answer 1

您可以使用DATE_ADD()：

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

Answer 2

您可以将其按使用的日期列进行分组，如果需要多天，则可以添加dateofbooking >= some_start_date和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";

Answer 3

您可以使用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。

每天最多5天如何获取总计数值

3 个答案: