尝试更新记录，但在php中失败

Question

im试图通过$_GET获取帖子的id来更新记录，然后通过$_post更新记录。

我已经通过$_GET执行了删除操作，它也很好用，mysqli_fetch_assoc也很好地显示要编辑的记录，但实际的编辑没有发生——它在空检查函数中的代码验证中给出了一个空错误。

我已经做了很多研究，但似乎无法理解这个错误，如果有人能建议对代码进行任何更改，我将非常感谢。

提前谢谢！

这是错误

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

以下是代码

<?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;
}
}
?>

这是html部分

<div  class="panel-body">
<form action="edit.php" method="POST">
<div class="form-group">
<label for="continent_name">Continent Name</label>
<input required type="text" placeholder="10" class="form-control" value="<?php echo  $c_name ; ?>" name="continent_name">
</div>
<small>Date Added: <?php echo  $c_dated_added ; ?></small> / <small>Last Edited: <?php echo  $c_last_edited ; ?></small>
<div class="form-group">
<input class="form-control btn-success" type="submit" name="edit" value="Submit">
</div>
</form>
</div>

Thid the while loop

<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>

Answer 1

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当你提交表单时，GET参数丢失了。你需要在表单的动作属性中传递它：

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

也请看一下这篇文章中的建议：有没有一种方法可以同时使用GET和POST？