duanhemou9834 2016-04-23 13:54
浏览 135
已采纳

获取数据并提交而不刷新页面

 ////////////////// Jquery AJAX submit zonder te refreshen ///////////////////////////

           function submitdata() {
              var vakken  = document.getElementById("vakken").value;
               var dataString = 'vakken=' + vakken;

                   // AJAX
                   $.ajax({
                       type: "POST",
                       url: "prototype.php",
                       data: dataString,
                       cache: false,
                       success: function(html) {
                           alert(html);
                       }

                   });

               return false;

           }

what is wrong here? //////////////////////////////////////////////////////////////////////////////////

<form id="selector" action="prototype.php" method="post" enctype="multipart/form-data">
                        <select name="vakken">

                            <option value="DED">VAK: DED  </option>
                            <option value="UXU">VAK: UXU  </option>
                            <option value="SCO">VAK: SCO  </option>
                            <option value="PO">VAK: PO  </option>
     </select>
                        <button type="button" onClick="submitdata();">Submitii</button>
                    </form>
                </div>

                <div id="vlakkencontainer">
                    <?php
    // recieve data 
                    if(isset($_POST['vakken'])) {
                        $vak = $_POST['vakken'];
                        echo $vak;
                        $sqldedquery = mysqli_query($con, "SELECT * FROM `opdracht` WHERE `vak` = '" . $vak . "'");
                        $vlakid = 0;
                        while ($row = mysqli_fetch_array($sqldedquery)) {

                            $afbeelding = $row['afbeeldingnaam'];
                            $pdf = $row['pdfnaam'];
                            $naamopdracht = $row['opdrachtnaam'];

                           echo '<a id="vakhover" href="../../../db/' . "$pdf" . '"><div class="vlak" id=' . "$vlakid++" . '>
            <img src="../../../db/' . $afbeelding . '"><div id="naamopdracht">' . $naamopdracht . '</div>
        </div></a>';
                        } 
                    } ?> 

how can i get this data out of my db without refresh the page. i tried ajax but i could not get it to work!

  • 写回答

3条回答 默认 最新

      报告相同问题?

      相关推荐 更多相似问题

      悬赏问题

      • ¥15 R语言,单因素cox检验,时间分层后,使用coz.zph()函数再次ph假设检验时报错,如何解决?
      • ¥15 如何预处理存在负值的样本数据,使其能够全都成为正的
      • ¥15 SW画图拖影,平滑处理如何关闭
      • ¥15 请问怎么通过css改变图片颜色
      • ¥15 c语言文件读取到结构体以及写入
      • ¥15 Blender: auto rig pro骨骼动画导出后变形穿模
      • ¥15 C51单片机的设计思路哈
      • ¥15 Linux脏牛提权漏洞
      • ¥15 为何我用uni-data-checkbox 标签在APP里调试无法显示?
      • ¥30 关于Stata软件OLS模型一些简单问题