PHP code，以排除使用index.php的水珠 [英] PHP code to exclude index.php using glob

问题描述

我想显示来自一个名为../health/文件随机页面
在这个文件中有一个index.php文件，并命名为118 PHP文件的其他文件。
我想随机显示从卫生文件夹中的文件，但我想它排除index.php文件。

I am trying to display a random page from a file called ../health/ In this file there is a index.php file and 118 other files named php files. I would like to randomly display a file from the health folder but i would like it to exclude the index.php file.

这下code包含index.php文件的时候。
我也试图改变$排除线来表示../health/index.php但仍没有运气。

This following code includes the index.php file sometimes. I have also tried altering the $exclude line to show ../health/index.php but still no luck.

<?php
$exclude = array("index.php"); // can add more here later
$answer = array_diff(glob("../health/*.php"),$exclude);
$whatanswer = $answer[mt_rand(0, count($answer) -1)];
include ($whatanswer);
?

另外code我曾尝试如下

Another code i have tried is the following

<?php
$exclude = array("../health/index.php"); // can add more here later
$health = glob("../health/*.php");
foreach ($health as $key => $filename) {
foreach ($exclude as $x) {
if (strstr($filename, $x)) {
unset($whathealth[$key]);
}
}
}
$whathealth = $health[mt_rand(0, count($health) -1)];
include ($whathealth);
?>

这code还包括index.php文件，但而不是显示的页面将显示页面错误。

This code also includes the index.php file but rather than showing the page it displays the page as an error.

推荐答案

这来考虑的第一件事是 array_filter（），居然有人 preg_grep（），但是这并不重要：

The first thing that came to mind is array_filter(), actually it was preg_grep(), but that doesn't matter:

$health = array_filter(glob("../health/*.php"), function($v) {
    return false === strpos($v, 'index.php');
});

使用 preg_grep（）使用 preG_GREP_INVERT 排除模式：

$health = preg_grep('/index\.php$/', glob('../health/*.php'), PREG_GREP_INVERT);

的它避免了使用回调，虽然实际上它很可能有同样的表现的

更新

满code应为您的特定情况下工作：

The full code that should work for your particular case:

$health = preg_grep('/index\.php$/', glob('../health/*.php'), PREG_GREP_INVERT);
$whathealth = $health[mt_rand(0, count($health) -1)];
include ($whathealth);

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