如何在ajax中发送表单结果

Question

您好我选择框并提交按钮并依赖于我想将表单结果发送到ajax的提交并检索数组结果。 这是我的PHP代码：

   if($param['aktion'] == 'get-widget-news-edit')
    {        


        $html = '<form name="UserInformationForm" method="POST" >
                 <table id="news"> ';
                  if(($post['news'])==4){

                $sql=" SELECT DISTINCT ad_news_texte.headline, ad_news.datum_archiv
            FROM ad_news_texte
            INNER JOIN ad_news_oe ON ad_news_texte.news_id = ad_news_oe.id_ad_news
            INNER JOIN ad_news ON ad_news_oe.id_ad_news = ad_news.id
            WHERE ad_news.datum_archiv
            BETWEEN curdate( ) - INTERVAL DAYOFWEEK( curdate( ) ) +28
            DAY AND curdate( )
            ";

                $sql_select=mysql_query($sql);
                while($row = mysql_fetch_array($sql_select)) {
                    echo $row['headline'] . " " .$row['datum_archiv'] ;
                    echo "<br>";
                }
            }

$html .= '<select name="news" id="news">
          <option value="4" '. (($_POST['news']=="4")  ?  "selected=selected" : "" ) .'>Show news from last 4 weeks</option>
           </select>                    
           </table>
           '.CreateButton($page['button']).'
           </form> 
          <div class="form_result"> </div>';

        $return = array(
                'status' => 1,
                'html'  => $html
            );

        echo json_encode($return);
        die();
    }

Js代码将选择框数组结果发送到ajax但无法执行此操作..

function saveNewsWidget()
    {
        var selectBoxValue = $('#UserInformationForm').find('.form_result').html();
        $.ajax({
        type: "POST",
        url: "ajax/dashboard.php",
        dataType : 'json',
        cache: false,
        data: {'aktion' : 'save-widget-news', 'news' : selectBoxValue },
        success: function(data)
        {
            //$('#news').html(data.html);
            getNewsWidget();
            //getNewsWidget();
            $('#news').html(data['html']);      
        }
      });
  }