Question

我是php html的新手。在我的应用程序中，我有一个按钮和href。我想制作按钮点击弹出窗口和3个单选按钮。我想使用无线电开关更新数据库中的数据而无需任何提交按钮。但我无法打开popover.href用于模态打开和查看数据。

我想当按钮点击弹出窗口时，单选按钮数据进入mysql数据库，但我无法做到。我想要模态和弹出。我怎么能这样做？

$('#exampleModal').on('show.bs.modal', function(event) {
  var button = $(event.relatedTarget) // Button that triggered the modal
  var recipient = button.data('whatever') // Extract info from data-* attributes
  var modal = $(this);
  var dataString = 'bclassid=' + recipient;
  $.ajax({
    type: "GET",
    url: "editdata.php",
    data: dataString,
    cache: false,
    success: function(data) {
      console.log(data);
      modal.find('.dash').html(data);
    },
    error: function(err) {
      console.log(err);
    }
  });
})

<div class="dataTable_wrapper">
  <table class="table table-striped table-bordered table-hover" id="dataTables-example">
    <thead>
      <tr style="background-color: #d7ccc8;">

        <th>CLASS</th>

        <th>OPTION</th>
      </tr>
    </thead>
    <tbody>
      <?php
        $result=mysqli_query($conn,"select bclassid,bclassname,status from bclassinfo");
        while($row = mysqli_fetch_array($result)) {
          echo "<tr class='odd gradeX'>";
          echo "<td align=center><a href='#myModal' data-whatever=" . $row[0] ."  data-toggle='modal' data-target='#exampleModal '>" .  $row[1] . "</a></td>";
          echo "<td><a class='btn btn-small btn-primary'">Edit</a></td>";//button to open popover with three radio button
          echo "</tr>";
            }
          ?>
    </tbody>
  </table>
</div>


<!-- Custom Theme JavaScript -->
<script src="../../dist/js/sb-admin-2.js"></script>
<script src="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/js/bootstrap.min.js" integrity="sha384-Tc5IQib027qvyjSMfHjOMaLkfuWVxZxUPnCJA7l2mCWNIpG9mGCD8wGNIcPD7Txa" crossorigin="anonymous"></script>

如何通过按钮单击弹出窗口将单选按钮数据发送到数据库

0 个答案: