Question

我很困惑地使用我的表格中的第一行id来传递我的模态形式。

显示我的模态的表格和按钮

<div id="ManageAccounts" class="tab-pane fade">
<h2>List of Customers</h2>         
<table class="table table-hover">
<thead>
  <tr>
    <th>Costumer ID</th>
    <th>Name</th>
    <th>Address</th>
    <th>Phone Number</th>
  </tr>
</thead>
<tbody>
    <?php
    while ($row_costumer = mysqli_fetch_assoc($result_costumer)) {
        $user_edit = $row_costumer['id'];
        echo "<tr>";
        echo "<td>" . $row_costumer['id'] . "</td>";
        echo "<td>" . $row_costumer['name'] . "</td>";
        echo "<td>" . $row_costumer['address'] . "</td>";
        echo "<td>" . $row_costumer['phone_number'] . "</td>";
        echo "<td>" . "<a class='edit-user btn btn-danger' href='#EditUser' data-toggle='modal' data-id='.$user_edit'>" . "<span class='glyphicon glyphicon-edit'>" . "</span>" . " Edit" . "</a>" . "</td>";
        echo "<tr>";
      }
    ?>
  </tbody>
 </table>
</div>

模态

<div class="modal fade" id="EditUser" tabindex="-1" role="dialog" aria-
labelledby="exampleModalLongTitle" aria-hidden="true">
<div class="modal-dialog" role="document">
<div class="modal-content">
  <div class="modal-header">
    <button type="button" class="close" data-dismiss="modal" aria-
 label="Close">
      <span aria-hidden="true">&times;</span>
    </button>
    <h3 class="modal-title" id="exampleModalLongTitle">Update User</h3>
  </div>
  <div class="modal-body">
    <form class="registerform" action="../includes/user_update.php" 
 method="POST">
      <div class="form-group">
        <input type="text" class="form-control" id="uname" name="uname" 
required placeholder="Name">
      </div>
      <div class="form-group">
        <input type="text" class="form-control" id="uaddress"  
name="uaddress" required placeholder="Address" required>
      </div>
      <div class="form-group">
        <input type="text" class="form-control" id="up_number" 
name="up_number" required placeholder="Phone Number" required>
      </div>          
      <div class="form-group">     
          <button type='submit' class="btn btn-primary lg btn-block" 
name="submit">Save</button>
       </div>
      </form>
     </div>
   </div>
  </div>
 </div>

user_update.php

<?php
include '../includes/config.php';

if (isset($_POST['submit'])) {
$uname = $_POST['uname'];
$uaddress = $_POST['uaddress'];
$up_number = $_POST['up_number'];
$u_id = $_POST['id'];
}

$query = "UPDATE costumer_inf SET name = '$uname', address = '$uaddress', 
phone_number = '$up_number' WHERE id = '$u_id'";
$result = mysqli_query($conn, $query);
header("location: ../admin/");

?>

使用php将数据从表传递到模态

0 个答案: