如何遍历目录中的文件并在<div> [duplicate]中输出每个文件的内容

Question

Possible Duplicate:
Loop code for each file in a directory

I assumed that I can reach this with following code but i think i make someting wrong with this code include($file[$])

<?php
$dir= opendir(".") or die ('Cannot open directory');
$div="<div class=\"left1\">
";
echo "<div class=\"wrapper\">
";
for( $i=0; $i<6; $i++){
    while(($file = readdir($dir)) != false){
    if(preg_match("/php$/", $file)){
    echo $div;
    include($file[$i]); // HERE ITS NOT WORKING AS I WANT
    echo "</div>
";
    }
    }
}
echo "</div>
";

?>

Answer 1

$file is a plain string at this point, not an array. So instead of include($file[$i]), you merely need to include($file).

while(($file = readdir($dir)) !== false){
  if(preg_match("/php$/", $file)){
      echo $div;
      include($file);
      echo "</div>
";
   }
}

The entire outer for loop appears to be unnecessary.