读出并以列表形式显示txt文件的内容

Question

我有一个脚本，列出了文件夹中的所有txt文件。我想要做的是显示文件内容而不是文件名。但我不得逞......我ReadFile的尝试和filegetcontents但我只得到.TXT命名

<?php 
if ($handle = opendir('../plugins/guestbook/news/')) { 
    while (false !== ($file = readdir($handle))) { 
     if ($file != "." && $file != "..") { 
      echo "<center><label class='tablog3a'>$file <div style='float: right;margin-top:10px;'><a href='gbremove.php?file=".$file."'><img src='images/deletegb.png' title='Verwijder afbeelding' class='clickreverse'></a></div></label></center>"; 
     } 
    } 
    closedir($handle); 
} 
?> 

我也尝试

<?php 
if ($handle = opendir('../plugins/guestbook/news/')) { 
    while (false !== ($file = readdir($handle))) { 
     if ($file != "." && $file != "..") { 
$txtfile = file_get_contents('$file'); 
      echo "<center><label class='tablog3a'>$txtfile<div style='float: right;margin-top:10px;'><a href='gbremove.php?file=".$file."'><img src='images/deletegb.png' title='Verwijder afbeelding' class='clickreverse'></a></div></label></center>"; 
     } 
    } 
    closedir($handle); 
} 
?> 

Answer 1

这个工作对我来说：

<?php 
$dir = "your_directory/"; // Your directory 

if ($handle = opendir($dir)) { 
    while (false != ($file = readdir($handle))) { 
     // Check for .txt files 
     if ($file != "." && $file != ".." && substr($file, -3) == "txt") { 
     // Open file to read 
     $fp = fopen($dir.$file, 'r'); 

     // Print contents 
     echo fread($fp, filesize($dir.$file)).'<br />'; 
     } 
    } 
    closedir($handle); 
}?> 

欢呼