php 根据星期几包含文件 - 简洁的代码？

Question

我已经使用之前关于堆栈溢出的一些问题的答案（感谢大家）提出了一个关联数组，其中包含基于星期几的特定文件，但我想知道是否有当路径名看起来很长且重复时，如何使代码更简洁？

$今天=日期("l");

$内容=数组（

"Sunday" => "/home/josull05/htdocs/wdd4/php/main-content/sundays-activities-home.php", "Monday" => "/home/josull05/htdocs/wdd4/php/main-content/mondays-activities-home.php", "Tuesday" => "/home/josull05/htdocs/wdd4/php/main-content/thursdays-activities-home.php", "Wednesday" => "/home/josull05/htdocs/wdd4/php/main-content/wednesdays-activities-home.php", "Thursday" => "/home/josull05/htdocs/wdd4/php/main-content/thursdays-activities-home.php", "Friday" => "/home/josull05/htdocs/wdd4/php/main-content/fridays-activities-home.php", "Saturday" => "/home/josull05/htdocs/wdd4/php/main-content/saturdays-activities-home.php"

); 包括 $content[$today];

Answer 1

您可以尝试以下代码段。

$today = date( 'l' );
$file = '/home/josull05/htdocs/wdd4/php/main-content/' . strtolower( $today ) . 's-activities-home.php';
include( $file );

变量 $today 将获得像 Sunday、Monday 这样的日期名称。在文件名中我们需要小写日期名称，所以使用strtolower。

注意：在包含之前检查文件是否存在是安全的。您可以使用 file_exists.

php 根据星期几包含文件 - 简洁的代码？

php include file based on day of the week - concise code?

arrays

date

include

days

dayofweek