如何在views.py中使用ajax请求发送好友请求而无需重新加载页面

from django.http import JsonResponse

def send_friend_request(request):
    if request.method == 'POST' and request.is_ajax():
        sender_id = request.POST.get('sender_id')
        receiver_id = request.POST.get('receiver_id')

        # 验证和权限控制逻辑
        # ...

        # 创建好友请求对象并保存到数据库
        # ...

        return JsonResponse({'success': True})
    else:
        return JsonResponse({'success': False, 'error': 'Invalid request'})

$(document).ready(function() {
    $('#send-friend-request-btn').click(function() {
        var sender_id = $('#sender-id').val();
        var receiver_id = $('#receiver-id').val();

        $.ajax({
            url: '/send_friend_request/',
            type: 'POST',
            data: {
                'sender_id': sender_id,
                'receiver_id': receiver_id
            },
            success: function(response) {
                if (response.success) {
                    // 请求成功处理逻辑
                    // ...
                } else {
                    // 请求失败处理逻辑
                    // ...
                }
            },
            error: function(xhr, status, error) {
                // 请求错误处理逻辑
                // ...
            }
        });
    });
});