3

下面使用的代码将从 GLOB_BRACE 中列出的 3 个文件夹中的任何一个中获取 10 个随机文件。

例如：

$files = (glob('../{folder1,folder2,folder3}/*.php', GLOB_BRACE));

我想在 $thelist 下面看到的 url 中回显文件夹名称

$thelist .= '<p><a href="../'.$folder 1 or 2 or 3.'/'.$file.'">'.$title.'</a></p>';

因此，当它显示在我的页面上时，它会显示。

<p><a href="../folder1/page-name.php">what ever</a></p>
<p><a href="../folder3/page-name.php">what ever</a></p>
<p><a href="../folder1/page-name.php">what ever</a></p>
<p><a href="../folder2/page-name.php">what ever</a></p>
<p><a href="../folder1/page-name.php">what ever</a></p>
<p><a href="../folder3/page-name.php">what ever</a></p>
<p><a href="../folder2/page-name.php">what ever</a></p>
<p><a href="../folder3/page-name.php">what ever</a></p>
<p><a href="../folder1/page-name.php">what ever</a></p>
<p><a href="../folder2/page-name.php">what ever</a></p>

使用的代码：

<?php 
$files = (glob('../{folder1,folder2,folder3}/*.php', GLOB_BRACE)); /* change php to the file you require either html php jpg png. */
shuffle($files);
$selection = array_slice($files, 0, 11);

foreach ($selection as $file) {
    $file = basename($file);
    if ($file == 'index.php') continue;

    $title = str_replace('-', ' ', pathinfo($file, PATHINFO_FILENAME));
        $randomlist .= '<p><a href="../'.$folder 1 or 2 or 3.'/'.$file.'">'.$title.'</a></p>';
    }
?>
<?=$randomlist?>
4

1 回答 1

1

将glob()返回目录和文件名。因此，如果您不重新分配$file到basename($file)，则整个字符串将保持不变以供输出。您仍然可以将条件签basename()入到。if()continue

foreach ($selection as $file) { 
  // Call basename() in the if condition, but don't reassign the variable $file
  if (basename($file) == 'index.php') continue;

  $title = str_replace('-', ' ', pathinfo($file, PATHINFO_FILENAME));
  // Using the full `$file` in the HTML output. No need for basename() or dirname().
  // Using htmlentities to encode the file path for an HTML attribute
  $randomlist .= '<p><a href="' . htmlentities($file, ENT_QUOTES) . '">'.$title.'</a></p>';
}
于 2013-01-20T22:28:35.200 回答