无需太多复制/粘贴即可很好地重定向错误？

Question

如果我有一个处理朋友管理的视图，这意味着有一个视图来处理添加、删除、阻止、取消阻止以及接受/拒绝成为朋友的邀请。我遇到的问题是，当我尝试向用户提供有意义的错误时，这些错误最终会出现在他们不应该出现的 url 上。

例如，如果 User1 和 User2 已经是朋友，并且 User1 转到将 User2 添加为的 url一个朋友而不是表单，就像他们不是朋友一样呈现，并且表单在 unique_together = (('user_from', 'user_to'),) 上失败，它会显示一条警告消息，并在显示表单之前将它们重定向到适当的页面。

这样

def add_friend(request, username):
    try:
        user = User.objects.get(username=username)
    except User.DoesNotExist:
        messages.error(request, 'A user with the username %s does not exist. \
            Try searching for the user below.' % username)
        return HttpResponseRedirect(reverse('friends_find_friend'))

    if Friend.objects.are_friends(request.user, user):
        messages.error(request, 'You are already friends with %s' % user)
        return HttpResponseRedirect(reverse('profiles_profile_detail', args=[user]))

如果没有这样的用户，这还包括检查和有意义的错误消息（而不是 404）。

这很容易处理，但通过其他检查它会增长到

def add_friend(request, username):
    try:
        user = User.objects.get(username=username)
    except User.DoesNotExist:
        messages.error(request, 'A user with the username %s does not exist. \
            Try searching for the user below.' % username)
        return HttpResponseRedirect(reverse('friends_find_friend'))

    if user == request.user:
        messages.error(request, 'You are already friends with yourself')
        return HttpResponseRedirect(reverse('friends_find_friend'))

    if Enemy.objects.is_blocked(request.user, user):
        messages.error(request, '%s has blocked you from adding them as a friend' % user)
        return HttpResponseRedirect(reverse('friends_find_friend'))

    if Enemy.objects.has_blocked(request.user, user):
        messages.error(request, 'You have blocked %s so you cannot add them as a friend' % user)
        return HttpResponseRedirect(reverse('profiles_profile_detail', args=[user]))

    if Friend.objects.are_friends(request.user, user):
        messages.error(request, 'You are already friends with %s' % user)
        return HttpResponseRedirect(reverse('profiles_profile_detail', args=[user]))

    if FriendRequest.objects.invitation_sent(request.user, user):
        messages.error(request, 'You already sent %s a request. You need to \
            wait for them to reply to it.' % user)
        return HttpResponseRedirect(reverse('friends_pending'))

    if FriendRequest.objects.invitation_received(request.user, user):
        messages.error(request, '%s already sent you a request and is waiting \
            for you to respond to them.' % user)
        return HttpResponseRedirect(reverse('friends_pending'))

所有这些都在

中再次复制

• remove_friend
• block_user
• unblock_user
• pending_invitations

如果视图在外壳中被绕过并且表单被独立使用，则会进一步复制为表单验证错误。

我要问的是，是否有一种更 Pythonic 的方式来完成此任务而无需过多的复制和粘贴？

编辑：

我一直在尝试解决这个问题，并且想知道这样的事情是否是一个好方法。

tests = ((Enemy.objects.is_blocked, 'This user has blocked you', reverse('friends_find_friend')),
         (Enemy.objects.has_blocked, 'You have blocked this user', reverse('profiles_profile_detail', args=[user])),)

for test in tests:
    if test[0](request.user, user):
        messages.error(request, test[1])
        return HttpResponseRedirect(test[2])

测试将在另一个类似于 url 模式的文件中定义，装饰器将包装视图函数以运行所有测试，如果有任何失败则重定向，如果一切正常，最后将其传递给视图函数。这会是一种有效的管理方法，不会用数百行样板代码阻塞视图文件吗？

编辑2：

我也有兴趣了解其他人如何做类似的事情。我怀疑我是第一个想要向用户显示消息而不是仅仅抛出 404 页面的人。

Answer 1

Python 装饰器非常适合。

定义你在装饰器中所做的所有验证，并用它装饰你所有的add_friend、remove_friend 等。

更新：

对于你的情况，它看起来像这样：

def do_friending_validation(fun):
    def validate_function(*args,**kwargs):
        try:
            user = User.objects.get(username=username)
        except User.DoesNotExist:
            messages.error(request, 'A user with the username %s does not exist. \
                Try searching for the user below.' % username)
            return HttpResponseRedirect(reverse('friends_find_friend'))

        if user == request.user:
        messages.error(request, 'You are already friends with yourself')
        return HttpResponseRedirect(reverse('friends_find_friend'))

        if Enemy.objects.is_blocked(request.user, user):
        messages.error(request, '%s has blocked you from adding them as a friend' % user)
        return HttpResponseRedirect(reverse('friends_find_friend'))

        if Enemy.objects.has_blocked(request.user, user):
        messages.error(request, 'You have blocked %s so you cannot add them as a friend' % user)
        return HttpResponseRedirect(reverse('profiles_profile_detail', args=[user]))

        if Friend.objects.are_friends(request.user, user):
        messages.error(request, 'You are already friends with %s' % user)
        return HttpResponseRedirect(reverse('profiles_profile_detail', args=[user]))

        if FriendRequest.objects.invitation_sent(request.user, user):
        messages.error(request, 'You already sent %s a request. You need to \
            wait for them to reply to it.' % user)
        return HttpResponseRedirect(reverse('friends_pending'))

        if FriendRequest.objects.invitation_received(request.user, user):
        messages.error(request, '%s already sent you a request and is waiting \
            for you to respond to them.' % user)
        return HttpResponseRedirect(reverse('friends_pending'))
        fun(*args,**kwargs)

    return validate_function

@do_friending_validation
def add_friend(request, username):

    #Do your stuff here

@do_friending_validation    
def remove_friend(request, username):

    #Do your stuff here

Answer 2

我想代码在 remove_friend 和 block_user 等内部并不完全重复（因为消息会有所不同） - 您可能可以将您的一些函数重构为单独的辅助函数，这些函数需要两个用户并返回某种指示成功或失败的状态。

如果没有您的pending_invitations 视图示例（这似乎与add_friend 最不同），很难提供更多帮助。