From https://stackoverflow.com/questions/14995307/sending-var-dump-to-firebug-console?

php_debug.php

function var2console($var, $name='', $now=false)
{
   if ($var === null)          $type = 'NULL';
   else if (is_bool    ($var)) $type = 'BOOL';
   else if (is_string  ($var)) $type = 'STRING['.strlen($var).']';
   else if (is_int     ($var)) $type = 'INT';
   else if (is_float   ($var)) $type = 'FLOAT';
   else if (is_array   ($var)) $type = 'ARRAY['.count($var).']';
   else if (is_object  ($var)) $type = 'OBJECT';
   else if (is_resource($var)) $type = 'RESOURCE';
   else                        $type = '???';
   if (strlen($name)) {
      str2console("$type $name = ".var_export($var, true).';', $now);
   } else {
      str2console("$type = "      .var_export($var, true).';', $now);
   }
}

function str2console($str, $now=false)
{
   if ($now) {
      echo "<script type='text/javascript'>\n";
      echo "//<![CDATA[\n";
      echo "console.log(", json_encode($str), ");\n";
      echo "//]]>\n";
      echo "</script>";
   } else {
      register_shutdown_function('str2console', $str, true);
   }
}