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eval() function not working in while loop


andytan91

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Hello guys i used the eval function but it does not echo anything back in the while loop ..the $new_string function contains "preg_split("/\.\(.*\)\.\(.*\)/", $file);", same as $parts..i need some guidance, thanks!

 

<?php


  //"$DOCUMENT_ROOT"."new/"; 
  $current_dir = 'C:\xampp\htdocs\Audit_Reports';   //Put in second part, the directory - without a leading slash but with a trailing slash!
  $dir = opendir($current_dir);        // Open the sucker
  

$newArgument = $_POST[argument];
$newArgument1 = $_POST[argument1];
  
$string ='preg_split("/\.\(.*\)\.\(.*\)/", $file);';
$new_string = preg_replace("#\/\\\.\\\\\(\.\*\\\\\)#","/\.\\($newArgument\)",$string);
$new_string = preg_replace("#\\\.\\\\\(\.\*\\\\\)#","\.\\($newArgument1\)", $new_string);


       
  echo ("<p><h1>List of Audit Reports:</h1></p><hr><br />");
  while ($file = readdir($dir))            // while loop
  {
  
  $parts = preg_split("/\.\(CLIENT\)\.\(.*\)/", $file);
  // $part = eval($new_string);

  
        if (is_array($parts) && count($parts) > 1) {    // does the dissected array have more than one part
        $extension = reset($parts);        // set to we can see last file extension
        if ($extension == "Audit_Report" OR $extension == "audit_report")    // is extension ext or EXT ?
             echo "<a href=\"$file\" target=\"_blank\"> $file </a><br />";    // If so, echo it out else do nothing cos it's not what we want
        }
    }
  echo "<hr><br />";
  closedir($dir);        // Close the directory after we are done


?>

 

 

 

The below codes are done in a while loop

 

Result:ArrayArrayArrayArrayArrayArrayArrayArrayArrayArra yArrayArray

$parts = preg_split("/\.\(A\)\.\(A\)/", $file);

  echo $parts;

 

Result:No Output

$part = eval($new_string);

echo $part;

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well the first thing I see wrong is

 

$string ='preg_split("/\.\(.*\)\.\(.*\)/", $file);';

 

You are trying to use a variable inside single quotes.  Single quotes do not parse variables, so you are assigning the literal string $file not the value of that variable.  You either need to put it inside double quotes or else break out of your single quotes and concatenate

 

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