Python 使用flask问题为搜索结果分页

我有个问题。我正在使用flask创建搜索结果。它工作得很好，但是当我尝试的时候 使用“页面导航”导航到下一页，它显示错误：

'TypeError:view函数未返回有效响应。函数要么返回None，要么在没有返回语句的情况下结束。“我做错了什么？”？代码如下所示：

@main.route("/filter", methods=['GET', 'POST'])
def filter():  

    if request.method == 'GET':
        #page = request.args.get('page', 1, type=int) 

        search_value = request.args.get('search')
        filter0 = "%{0}%".format(search_value)
        filter1 = request.args.get('myselect')
        filter2 = request.args.get('myselect2')
        if filter1 == 'new':

            if filter2 == "All":
                page = request.args.get('page', 1, type=int) 
                posts = Post.query.filter(Post.course_name.like(filter0)).filter_by().order_by(Post.timestamp.desc()).paginate(page=page, per_page=10)
            return render_template('home.html', posts=posts, filter1 = filter1, filter2 = filter2)

{posts.iter\u页面中的页数百分比（左页边=1，右页边=1，左页边=1，右页边=2）%}
{%if page_num%}
{%if posts.page==page_num%}
{%else%}
{%endif%}
{%else%}
...
{%endif%}
{%endfor%}

如果过滤器1！='新建'

，查看函数不返回任何内容

下面的代码可能会解决这个问题

@main.route("/filter", methods=['GET', 'POST'])
def filter():  

    if request.method == 'GET':
        pass
        if filter1 == 'new':
            if filter2 == "All":
               pass
            return render_template('home.html', posts=posts, filter1 = filter1, filter2 = filter2)
    return 'something'

@main.route("/filter", methods=['GET', 'POST'])
def filter():  

    if request.method == 'GET':
        pass
        if filter1 == 'new':
            if filter2 == "All":
               pass
            return render_template('home.html', posts=posts, filter1 = filter1, filter2 = filter2)
    return 'something'