如何使用PHP将目录中的文件名作为选项填充到下拉菜单？

问题描述：

I'm trying to create a drop down menu that points to a directory and populates a drop down menu with the names of certain files in that directory using PHP.

Here's what I'm working with:

<?php

$path = "pages/"; //change this if the script is in a different dir that the files you want
$show = array( '.php', '.html' ); //Type of files to show

$select = "<select name=\"content\" id=\"content\">";

$dh = @opendir( $path );

while( false !== ( $file = readdir( $dh ) ) ){
    $ext=substr($file,-4,4);
        if(in_array( $ext, $show )){       
            $select .= "<option value='$path/$file'>$file</option>
";
    }
}  

$select .= "</select>";
closedir( $dh );

echo "$select";
?>

This bit of code is giving me an errors, and I'm not even really attached to it if there's a better way of trying to accomplish what I'm trying to do.

我正在尝试创建一个指向目录的下拉菜单，并使用名称填充下拉菜单使用PHP在该目录中的某些文件。 p>

以下是我正在使用的内容： p>

 ＆lt;？php 
 
  $ path =“pages /”;  //如果脚本位于您想要的文件的另一个目录中，则更改此项
 $ show = array（'。php'，'。html'）;  //要显示的文件类型
 
 $ select =“＆lt; select name = \”content \“id = \”content \“＆gt;”; 
 
 $ dh = @opendir（$ path）;  
 
而（假！==（$ file = readdir（$ dh）））{
 $ ext = substr（$ file，-4,4）; 
 if（in_array（$ ext，$ show））  {
 $ select。=“＆lt; option value ='$ path / $ file'＆gt; $ file＆lt; / option＆gt; 
”; 
} 
} 
 
 $ select。=“＆lt; / 选择＆gt;“; 
closedir（$ dh）; 
 
echo”$ select“; 
？＆gt;  
  code>  pre> 
 
 这段代码给了我一个错误，如果有更好的方法来尝试完成我的工作，我甚至都不会真正依赖它 试图做。 p> 
  div>

答

It would be easier to use glob() because it can handle wildcards.

// match all files that have either .html or .php extension
$file_matcher = realpath(dirname(__FILE__)) . '/../pages/*.{php,html}';

foreach( glob($file_matcher, GLOB_BRACE) as $file ) {
  $file_name = basename($file);
  $select .= "<option value='$file'>$file_name</option>
";
}

答

You need a full path reference (i.e. /var/www/pages/) instead of just "pages".

Also you might consider using DirectoryIterator object for easily getting to directroy information (if you are using PHP 5).

答

I don't know, which errors you get. But I think it won't work with the $show array because you're comparing the last 4 chars of the file with the contents of the array. Instead of $ext=substr($file,-4,4); you could write $ext=substr($file, strrpos( $file, ".")); which gives you the string from the position of the last occurance of ".".

Also I suggest for test reason that you omit the @ opening the directory because I think that the path cannot be found.

如何使用PHP将目录中的文件名作为选项填充到下拉菜单？

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