php单击按钮，弹出以显示详细信息

Question

当我单击每行表中的“视图”按钮时，如何显示详细信息数据(使用弹出)？

按钮码

echo '<button href="student.php?id="'.$row['mactrixNo'].'" onclick=document.getElementById("id02").style.display="block">View</button>';

弹出代码

<div id="id02" class="logform">
    <table border="1" class="tg w-100">
        <tr>
            <?php if(isset($_GET['id'])){
                $query='select *
                from student
                where mactrixNo="'.$_GET['id'].'"';
            }
            ?>

当弹出时我无法得到'id‘

Answer 1

引导程序提供了实现这一目标所需的HTML、CSS和JavaScript。这是v4 https://getbootstrap.com/docs/4.1/components/modal/寄来的

<!-- Button trigger modal -->
<button type="button" class="btn btn-primary" data-toggle="modal" data-target="#exampleModal">
  Launch demo modal
</button>

<!-- Modal -->
<div class="modal fade" id="exampleModal" tabindex="-1" role="dialog" aria-labelledby="exampleModalLabel" aria-hidden="true">
  <div class="modal-dialog" role="document">
    <div class="modal-content">
      <div class="modal-header">
        <h5 class="modal-title" id="exampleModalLabel">Modal title</h5>
        <button type="button" class="close" data-dismiss="modal" aria-label="Close">
          <span aria-hidden="true">&times;</span>
        </button>
      </div>
      <div class="modal-body">
        ...
      </div>
      <div class="modal-footer">
        <button type="button" class="btn btn-secondary" data-dismiss="modal">Close</button>
        <button type="button" class="btn btn-primary">Save changes</button>
      </div>
    </div>
  </div>
</div>

只要您在<head>部分中有CSS和JS，您所需要做的就是编辑内容！