Sendidng动态id和接收数据

Question

我是JS，AJAx的初学者，等等。我在互联网上搜索并发现了一个代码，所以我用它，稍微改了一下......但似乎它没有用（它什么也没做）。 首先是javascript和html：

＆＃13;

＆＃13;

function fetch_select(val)
{
   $.ajax({
 type: 'post',
 url: 'fetch_data.php',
 data: {
   get_option:val
 },
 success: function (response) {
   document.getElementById("new_select").innerHTML=response; 
 }

   });

}

＆＃13;

        <select onchange="fetch_select(this.value);">
           <option>Wybierz</option>';
		      $zapytanie = "
			    SELECT
				  scan_category.scat_id,
				  scan_category.long_name
			    FROM
				  scan_category
			    WHERE
				  scan_category.state = 1 OR scan_category.state = 4 and scan_category.hide = 0
    			ORDER BY
	              scan_category.long_name";
              $wykonaj = mysql_query($zapytanie) or die(mysql_error());           
               while($row=mysql_fetch_array($wykonaj))
               {
                 echo '<option>'.$row[1].'</option>';
               }
               echo'</select>
               <div id="new_select">
               </div>';

＆＃13;

＆＃13;

＆＃13;

     $manga = $_POST['get_option'];
     $znajdz = "
        SELECT users.nick, scan_category.long_name, scan_work.work_name, scan_roles.role_id, scan_roles.scat_id, scan_roles.uid, scan_roles.what_id
        FROM scan_roles
        INNER JOIN scan_work ON scan_roles.what_id = scan_work.work_id
        INNER JOIN users ON scan_roles.uid = users.uid
        INNER JOIN scan_category ON scan_category.scat_id = scan_roles.scat_id
        WHERE
        scan_roles.scat_id = '$manga'";
     $wykonaj = mysql_query($znajdz or die(mysql_error());
         while ($row = mysql_fetch_array($znajdz)) {
           echo '<h2>'.$row[1].
           '</h2><br />'.$row[2].
           ':
        <form enctype="multipart/form-data" action="manage.php5?mode=edit_role&edit&role_id="'.$row[4].
           ' method="POST">
          <input type="hidden" name="pass" value="'.$_POST['pass'].
           '">';
           echo('<select name="user_id" size="1">');
           $zapytanie = '
                SELECT
                  users.uid,
                  users.nick
                FROM
                  users
                WHERE
                  active > 0
                 ORDER BY
                  users.nick';
           $wykonaj2 = mysql_query($zapytanie) or die(mysql_error());
           while ($wiersz = mysql_fetch_array($wykonaj)) {
             echo('<option value="'.$wiersz[0].
               '"');
             if ($wiersz[0] == $row[6]) echo ' selected="selected"';
             echo('>'.$wiersz[1].
               '</option>');
           }
           echo '</select>';

           echo('<input type="submit" value="edit">
        </form>');

你知道如何让它发挥作用（它似乎甚至不会去文件＆＃34; fetch_data.php）。

如果我要添加其他内容，请告诉我，我几乎是初学者！

当我输入console.log（响应）时，它确实说ReferenceError：响应没有定义，当我把它打成这样 - console.log（＆＃34;响应＆＃34;）它给了我＆＃34;响应＆ ＃34;