Python 以10为基数的int()的文本无效:';aa';,,I';I’’我不想通过电话,为什么我会收到这个?
您好,我是django activity stream,当我尝试跟踪名为aa的类别时,遇到了以下错误:以10为基数的int()的文本无效。这是我的密码Python 以10为基数的int()的文本无效:';aa';,,I';I’’我不想通过电话,为什么我会收到这个?,python,django,Python,Django,您好,我是django activity stream,当我尝试跟踪名为aa的类别时,遇到了以下错误:以10为基数的int()的文本无效。这是我的密码 def category_timeline(request, category): user = User.objects.select_related('profile').get(category=category) user_actions = [] if is_following(request.user, us
def category_timeline(request, category):
user = User.objects.select_related('profile').get(category=category)
user_actions = []
if is_following(request.user, user) or not user.profile.private:
user_actions = actor_stream(user)
context = {
'user': user,
'activities': user_actions,
}
return render(request, 'timeline.html', context)
def follow_user(request, category):
follow(request.user, User.objects.get(category=category))
return redirect('category_timeline', category)
def unfollow_user(request, category):
unfollow(request.user, User.objects.get(category=category))
return redirect('category_timeline', category)
这是我的追踪
它发生在这里user=user.objects.select_related('profile').get(category=category)
我不确定我是否做对了。我试图做的是显示用户,以便能够跟踪类别,并显示类别的内容。
我有类别模型和类别视图
看起来像这样
def category(request, category_name_url):
user = User.objects.get(username=request.user)
category_name = decode_url(category_name_url)
category = Category.objects.get(name=category_name)
回溯:
File "env/local/lib/python2.7/site-packages/django/core/handlers/base.py" in get_response
132. response = wrapped_callback(request, *callback_args, **callback_kwargs)
File "rclone/main/views.py" in follow_user
271. follow(request.user, User.objects.get(category=category))
File "env/local/lib/python2.7/site-packages/django/db/models/manager.py" in manager_method
127. return getattr(self.get_queryset(), name)(*args, **kwargs)
File "env/local/lib/python2.7/site-packages/django/db/models/query.py" in get
325. clone = self.filter(*args, **kwargs)
File "env/local/lib/python2.7/site-packages/django/db/models/query.py" in filter
679. return self._filter_or_exclude(False, *args, **kwargs)
File "env/local/lib/python2.7/site-packages/django/db/models/query.py" in _filter_or_exclude
697. clone.query.add_q(Q(*args, **kwargs))
File "env/local/lib/python2.7/site-packages/django/db/models/sql/query.py" in add_q
1309. clause, require_inner = self._add_q(where_part, self.used_aliases)
File "env/local/lib/python2.7/site-packages/django/db/models/sql/query.py" in _add_q
1337. allow_joins=allow_joins, split_subq=split_subq,
File "env/local/lib/python2.7/site-packages/django/db/models/sql/query.py" in build_filter
1199. lookups, value)
File "env/local/lib/python2.7/site-packages/django/db/models/fields/related.py" in get_lookup_constraint
1751. lookup_class(target.get_col(alias, source), val), AND)
File "env/local/lib/python2.7/site-packages/django/db/models/lookups.py" in __init__
101. self.rhs = self.get_prep_lookup()
File "env/local/lib/python2.7/site-packages/django/db/models/lookups.py" in get_prep_lookup
139. return self.lhs.output_field.get_prep_lookup(self.lookup_name, self.rhs)
File "env/local/lib/python2.7/site-packages/django/db/models/fields/__init__.py" in get_prep_lookup
727. return self.get_prep_value(value)
File "env/local/lib/python2.7/site-packages/django/db/models/fields/__init__.py" in get_prep_value
985. return int(value)
Exception Type: ValueError at /follow/
发布上述讨论的最终结论 在这个调用中,
User.objects.get(category=category))
应该是category模型的主键,而不是字符串
所以像下面这样的方法应该可以奏效
category_id = Category.objects.get(name=category)
User.objects.get(category=category_id))
看起来
profile
model中的一个字段定义为整数,数据库为此返回“aa”。@shryas,感谢您的建议,但我没有配置文件模型或任何定义为整数的模型字段…不确定此django活动流如何工作,但…我没有任何int字段,但我遇到了此错误…我说的是这一行User.objects。选择与之相关的('profile')。获取(category=category)
profile
是自定义模型的代理名称,用于扩展User
模型。你能验证这是哪个模型吗?谢谢你的回复,我正在关注django活动流,只是盲目地关注代码,没有意识到“profile”需要是category,因为我有用户需要遵循的category模型。但即使在我切换到类别之后,我也会遇到同样的错误…我只是尝试使用活动流,但这对我来说似乎太超前了…你能帮我一下吗?你能告诉我你正在跟踪的文档吗?