正在尝试更新记录，但在php中失败

我试图通过$\u GET获取帖子的id来更新记录，然后通过$\u post更新记录

我已经通过$\u GET执行了删除操作，并且mysqli\u fetch\u assoc在显示记录以进行编辑时也可以正常工作，但实际编辑没有发生。空检查函数中的代码验证中出现空错误

我已经做了很多研究，但我似乎无法理解这个错误，如果有人能建议修改代码，我将不胜感激

提前谢谢你

这就是错误所在

Notice: Undefined index: id in /then the long url etc/

下面是代码

<?php
 //DB Connection
  include'include/db-conn.php';
  if (isset($_POST['edit'])) {
     //Raw GET Inputs
    $raw_c_id       = $_GET['id'];

    //Cleaned Inputs
    $c_c_id         = filter_var($raw_c_id, FILTER_SANITIZE_STRING);

    //Error Mwssages
    $empty      = '<div class="alert alert-danger alert-dismissible">
                <a href="#" class="close" data-dismiss="alert" aria-label="close">&times;</a>
                <strong>Error!</strong>Field is empty please provide content!
              </div>
              ';
    $success    = '<div class="alert alert-success alert-dismissible fixed-top">
                <a href="#" class="close" data-dismiss="alert" aria-label="close">&times;</a>
                <strong>Success!</strong> Content Added Successfully
              </div>
              ';
    $not_success  = '<div class="alert alert-danger alert-dismissible">
                    <a href="#" class="close" data-dismiss="alert" aria-label="close">&times;</a>
                    <strong>Not Success!</strong> Content Not Added Successfully
                  </div>
                  ';

      if (empty($c_c_id)) {
        echo $empty;
        exit();
        header("Location:index.php");
      }

      $update      = "UPDATE `continents` 
       SET `continent_name`='$c_c_name', `last_edited`='date(d/m/Y)'  
       WHERE `id`='$c_c_id'";
       $run_update  = mysqli_query($conn, $update);
      if (!$run_update) {
        header("Location: index.php");
        echo $not_success;

      }
      else{
        header("Location: index.php");
        echo $success;
      }
  }
 ?>

---

看起来您正试图同时使用GET和POST参数。它不起作用的原因是提交表单时GET参数丢失。您需要在表单的action属性中传递它：

<form action="edit.php?id=<?php echo $_GET['id'] ?>" method="POST">

您可以发布它发布到的URL吗。我想看看GET Parameters如何显示您的实际url。非常感谢您的帮助，它确实起了作用。有没有机会从我这里看一下这个代码，让我知道我哪里弄错了！如果此答案解决了您的问题，请将其标记为正确答案。我会看看你的其他帖子。
<div class="table-responsive">
<table id="example" class="table table-hover ">
    <thead>
      <tr class="">
        <th>ID</th>
        <th>Continent Name</th>
        <th>Date Added</th>
        <th>Status</th>
        <th>Edit</th>
        <th>Delete</th>
      </tr>
    </thead>
    <tbody>
        <?php 
            $all_continents     =   "SELECT * FROM `continents` ORDER BY `status`";

            $run                    =   mysqli_query($conn,$all_continents);

            while ($row_result = mysqli_fetch_assoc($run)) {
                $id =   $row_result['id'];
                $c_continent_name = $row_result['continent_name'];
                $c_date_added = $row_result['date_added'];
                $c_status = $row_result['status'];

                echo "
                <tr>
                    <td>$id</td>
                    <td>$c_continent_name</td>
                    <td>$c_date_added</td>
                    <td>$c_status</td>
                    <td>
                        <a class='btn btn-info' href='edit.php?id=$id'>Edit</a>
                    </td>
                    <td>
                        <a class='btn btn-danger'  href='delete.php?id=$id'>Delete</a>
                    </td>
                </tr>
                ";
            }
         ?>
    </tbody>
</table>

<form action="edit.php?id=<?php echo $_GET['id'] ?>" method="POST">