在基于类的视图中使用PermissionRequiredMixin时,我试图返回404而不是403,我不希望用户看到403,因为这意味着URL存在
<?php
$category_ids = get_all_category_ids();
$current_category = get_category( get_query_var( 'cat' ) );
$cat_id = $current_category->cat_ID;
$args = array(
'orderby' => 'slug',
'parent' => $cat_id
);
$categories = get_categories( $args );
foreach ( $categories as $category ) {
echo '<div><h3 class="title">' . $category->name . '</h3><div class="text">' . $category->description . '</div><div class="ink"><a href="' . get_category_link( $category->term_id ) . '">Read more</a></div></div>';
}
?>
有什么想法吗?
答案 0 :(得分:1)
一种方法是定制403模板
project / urls.py
handler403 = 'your_app.views.handler403'
现在在403视图中,您可以呈现404模板。
def handler403 (request, *args, **kwargs):
return render(request, '404.html')