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Home/ Questions/Q 6960295
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Editorial Team
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Editorial Team
Asked: May 27, 20262026-05-27T15:23:03+00:00 2026-05-27T15:23:03+00:00

I have installed ConsoleExport Firebug extension. I have enabled auto-export feature at about:config by

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I have installed ConsoleExport Firebug extension. I have enabled auto-export feature at about:config by setting extensions.firebug.consoleexport.active to true. I have opened a page with JavaScript error, for example http://www.random.org. I have enabled Firebug, reloaded the page. I can see the error in log:

uncaught exception: Error: Permission denied for <https://www.facebook.com> to get property Proxy.InstallTrigger

The ConsoleExport page says:

You can also activate an auto-export feature that is sending
individual logs into specified server as XML packets.

I know I should set extensions.firebug.consoleexport.serverURL to server URL. I do not know how to catch XML packets that ConsoleExport sends.

For example, I know how to set up Apache on Mac, but what then? How to catch XML packets?

Environment:

  • Mac OS X 10.7.2
  • Firefox 8.0.1
  • Firebug 1.8.4
  • ConsoleExport 0.5b4

I have access to Windows and Linux machines, if the server that catches XML packets is easier to set up there.

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  1. Editorial Team
    Editorial Team
    2026-05-27T15:23:04+00:00Added an answer on May 27, 2026 at 3:23 pm

    Modified script fixing a problem with getting the posted data:

    Note that the content-Type of the posted data is application/xml

    <?php
      $filename = 'consoleexport.log';
    
      if (!$handle = fopen($filename, 'a'))
      {
        echo 'File "'.$filename.'" could not be opened';
        exit;
      }
    
      ob_start();
      $content = file_get_contents('php://input');
      $content .= "\n";
      ob_clean();
    
      if (!fwrite($handle, $content))
      {
        echo 'Can\'t write into file "'.$filename.'"';
        exit;
      }
    
      echo 'Done!';
      fclose($handle);
    ?>
    

    Honza

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