从目录中读取文件并回显某些行

I want to read all files from a directory but instead of displaying the first character i want to display a certain line, f.e. line 4.

    <?php
$directory = "content/";
$dir = opendir($directory);
while (($file = readdir($dir)) !== false) {
  $filename = $directory . $file;
  $type = filetype($filename);
  if ($type == 'file') {
     $contents = file_get_contents($filename);
     $items = explode("|", $contents);
     foreach ($items as $item) {
       echo "$item[0]";
     }
  }
}
closedir($dir);
?>

Thanks!

Could you show your file structure, please? Basicaly you can use something like this:

$contents = file_get_contents($filename);
$items = explode("
", $contents); //explode file content
foreach ($items as $item) {
   echo $item[3]; //echo your line
}

You can use file() function. It reads files into an array, where every line will be an array member, to skip empty lines pass the FILE_SKIP_EMPTY_LINES flag paramater.

For more info consult the docs.

// `$items` is already an array
$items = explode("|", $contents);
// if you want first element of array just:
echo $item[0];
// if you want fourth element of array just:
echo $item[3];
// without `foreach`