我正在制作一个菜谱应用程序，更新一个菜谱方法时遇到了问题。我有更新表单和sql查询。我甚至检查了表单输入变量 name , description ，和 url 在我的 if 语句是更新的值。他们是。。我的sql是有效的。然而，当我填写编辑表单并点击submit时，我被导航回菜谱列表页面，其中我编辑的菜谱没有更改。
这是我的密码。

<?php 

$recipe_id = $_GET['recipe_id'];
$recipes_fetch_existing_row_query = mysqli_query($con, "SELECT * FROM recipes WHERE id='$recipe_id'");
$recipe = mysqli_fetch_array($recipes_fetch_existing_row_query);
$name = $recipe['name'];
$description = $recipe['description'];
$url = $recipe['url'];

if(isset($_POST['update_recipe'])) {
    $name = $_POST['recipe_name'];
    $description= $_POST['recipe_description'];
    $url = $_POST['recipe_url'];
    $recipe_update_query = mysqli_query($con, "UPDATE recipe SET name='$name', description='$description', url='$url' WHERE id='$recipe_id'");
    header("Location: ../../../index.php");
}
?>

<form action="../../../index.php" class="update_recipe_form" method="POST">
  <input type="text" name="recipe_name" placeholder="name" value="<?php echo $recipe_name ?>" required>
  <textarea name="recipe_description" id="recipe_description" placeholder="Enter a brief description of the recipe"><?php echo $recipe_description; ?></textarea>
  <input type="text" name="recipe_url" id="recipe_url" value="<?php echo $recipe_url ?>" placeholder="url" required>
  <input type="submit" name="update_recipe" id="update_recipe" value="Update">
</form>

Any ideas? :(