透過 Cron Job 運行這個 PHP 文件

Question

這是腳本：

---

<?php
//Create back files?
define('CREATE_BACKUPS', FALSE);

if (!is_dir($argv[1]))
{
   echo "You must enter a valid path such as /home/apresv/public_html or apresv/public_html for this script to function.\n";
   exit;
}

//Search the path for all php files, opening each one, and checking to see if it's infected

//First, get an array list of all valid .php files.


$files = listdir($argv[1]);
foreach ($files as $filename)
{
   //We only need to check php files, so we add that here
   if (file_extension($filename) == 'php')
   {
      //This is a php file so lets check it to see if it's infected.
      $contents = file_get_contents($filename);
      $backup = $contents;

      //There will always be 2 opening tags in an infected file and since the hack is always at the top, it's easiest to test for this right away.
      $test = between('<?php', '<?php', $contents);

      //This particular hack likes to use toolbarqueries so we test to see if our chunk is an infected chunk.  If your website uses this url somehow, then add extra if statements as necessary.
      if (after('toolbarqueries', $test))
      {
         //This chunk is infected.  So lets replace it and resave the file.
         $contents = str_replace('<?php'.$test.'<?php', '<?php', $contents);

         //Now save it! Woohoo!
         file_put_contents($filename, $contents);
         if (CREATE_BACKUPS)
         {
            file_put_contents($filename.'.orig', $backup);
         }

         echo "$filename has been cleaned.\n";
      }
   }
}

function after ($this, $inthat)
    {
        if (!is_bool(strpos($inthat, $this)))
        return substr($inthat, strpos($inthat,$this)+strlen($this));
    };

    function after_last ($this, $inthat)
    {
        if (!is_bool(strrevpos($inthat, $this)))
        return substr($inthat, strrevpos($inthat, $this)+strlen($this));
    };

    function before ($this, $inthat)
    {
        return substr($inthat, 0, strpos($inthat, $this));
    };

    function before_last ($this, $inthat)
    {
        return substr($inthat, 0, strrevpos($inthat, $this));
    };

    function between ($this, $that, $inthat)
    {
     return before($that, after($this, $inthat));
    };

    function between_last ($this, $that, $inthat)
    {
     return after_last($this, before_last($that, $inthat));
    };

    // USES
    function strrevpos($instr, $needle)
    {
        $rev_pos = strpos (strrev($instr), strrev($needle));
        if ($rev_pos===false) return false;
        else return strlen($instr) - $rev_pos - strlen($needle);
    };

    function listdir($dir='.') {
    if (!is_dir($dir)) {
        return false;
    }

    $files = array();
    listdiraux($dir, $files);

    return $files;
}

function listdiraux($dir, &$files) {
    $handle = opendir($dir);
    while (($file = readdir($handle)) !== false) {
        if ($file == '.' || $file == '..') {
            continue;
        }
        $filepath = $dir == '.' ? $file : $dir . '/' . $file;
        if (is_link($filepath))
            continue;
        if (is_file($filepath))
            $files[] = $filepath;
        else if (is_dir($filepath))
            listdiraux($filepath, $files);
    }
    closedir($handle);
}

function file_extension($filename)
{
   $info = pathinfo($filename);
   return $info['extension'];
} 
?>

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當我嘗試透過 cron 執行此腳本時，我得到輸出「您必須輸入有效路徑，例如 /home/apresv/public_html 或 apresv/public_html 才能使該腳本運作。

我需要做什麼才能讓它通過 CRON 作業運行？

謝謝！

Answer 1

使用 CRON 執行腳本的基本方法是使文件可執行，談到文件時，php您必須為文件提供解釋器命令所在的路徑，這稱為shebang（了解更多信息，請參閱這裡）並且必須放置在腳本的最頂部。

所以你必須執行以下操作

$ chmod +x script.php
$ sed -i '1 i\#!/usr/bin/php' script.php #This is to insert the shebang

這樣就可以配置CRON了

$ crontab -e
* * * * * /path/to/script.php

筆記：當然，您必須確保您的腳本有效。