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Django:将 URL 参数传递给模板标签中反向 URL 查找的每个实例

[英]Django: passing a URL parameter to every instance of a reverse URL lookup in a template tag

Sorry if the title is unclear, I'm not sure the best way to describe the issue.抱歉,如果标题不清楚,我不确定描述问题的最佳方式。 I have an application with a Ticket model and a Team model.我有一个带有Ticket model 和Team model 的应用程序。 All Ticket s are associated with a single Team .所有Ticket都与一个Team相关联。 The issue I'm having is a problem of URL reversing.我遇到的问题是 URL 倒车的问题。 I'm trying to set it up my URLs like so: /<team_pk>/tickets/ displays a list of tickets associated with the Team specified by team_pk .我正在尝试像这样设置我的 URL: /<team_pk>/tickets/显示与team_pk指定的团队相关联的票证列表。 So /1/tickets/ would display all of the tickets for the first Team.所以/1/tickets/将显示第一个团队的所有门票。 Both of these objects are in the app tracker .这两个对象都在 app tracker中。

To do this, I've set up my project/urls.py files like so:为此,我设置了我的 project/urls.py 文件,如下所示:

project/urls.py项目/urls.py

urlpatterns = [ path('<team_pk>/', include('tracker.urls', namespace='tracker')), ]

tracker/urls.py跟踪器/urls.py

urlpatterns = [ path('tickets/', views.TicketTable.as_view(), name='ticket_list'), ]

Then, inside my html templates, I have the following URL tag:然后,在我的 html 模板中,我有以下 URL 标签:

href="{% url 'tracker:ticket_list' %}"

This results in a NoReverseMatch error:这会导致 NoReverseMatch 错误:

NoReverseMatch at /1/tickets/
Reverse for 'ticket_list' with no arguments not found. 1 pattern(s) tried: ['(?P<team_pk>[^/]+)/tickets/$']

What I would like is for the reverse match to just use the current value for the team_pk URL kwarg.我想要的是反向匹配只使用team_pk URL kwarg 的当前值。

What I have tried to fix it:我试图解决的问题:

I have found the following solution to the problem, but it involves a lot of repetition, and I feel like there must be a DRYer way.我找到了以下解决问题的方法,但是它涉及到很多重复,我觉得必须有一个 DRYer 的方式。

First, I extend the get_context_data() method for every view on the site.首先,我为站点上的每个视图扩展了get_context_data()方法。

def get_context_data(self, **kwargs):
    context = super().get_context_data(**kwargs)
    context['current_team_pk'] = self.kwargs['team_pk']
    return context

Then I reference this context in every URL template tag:然后我在每个 URL 模板标签中引用这个上下文:

href="{% url 'tracker:ticket_list' team_pk=current_team_pk %}"

This results in the desired behavior, but it's a lot of repetition.这会产生所需的行为,但需要大量重复。 So, is there a better way?那么,有没有更好的方法呢?

Edit:编辑:

Based on Willem Van Onsem's suggestion, I changed the URL template tag to href="{% url 'tracker:ticket_list' team_pk=team_pk %}" , referencing the URL kwarg directly. Based on Willem Van Onsem's suggestion, I changed the URL template tag to href="{% url 'tracker:ticket_list' team_pk=team_pk %}" , referencing the URL kwarg directly. But this doesn't seem to be working reliably.但这似乎并不可靠。 On the index page /<team_pk>/ loads just fine, and it includes two relevant URLs: /<team_pk>/ and /<team_pk>/tickets/ .在索引页面上/<team_pk>/加载得很好,它包括两个相关的 URL: /<team_pk>//<team_pk>/tickets/ When I navigate to /<team_pk>/tickets/ , however, I get another NoReverseMatch error:但是,当我导航到/<team_pk>/tickets/时,我收到另一个 NoReverseMatch 错误:

NoReverseMatch at /1/tickets/
Reverse for 'home' with keyword arguments '{'team_pk': ''}' not found. 1 pattern(s) tried: ['(?P<team_pk>[^/]+)/$']

It seems the URL kwarg <team_pk> is not being passed for some reason.似乎 URL kwarg <team_pk>由于某种原因没有通过。 But the only link to 'home' is part of my base.html , which the other templates are extending.但是到'home'的唯一链接是我的base.html的一部分,其他模板正在扩展。 So the relevant template tags are the same.所以相关的模板标签是一样的。

Edit 2:编辑2:

The view in question:有问题的观点:

class TicketTable(LoginRequiredMixin, SingleTableMixin, FilterView):
    table_class = my_tables.TicketTable
    template_name = 'tracker/ticket_list.html'
    filterset_class = TicketFilter
    context_object_name = 'page'
    table_pagination = {"per_page": 10}
    PAGE_TITLE = 'Open Tickets'
    TICKET_STATUS_TOGGLE = 'Show Closed Tickets'
    TICKET_STATUS_TOGGLE_URL = 'tracker:closed_ticket_list'
    DISPLAY_DEV_FILTER = True
    DISPLAY_STATUS_FILTER = True

    def get_context_data(self, **kwargs):
        context = super().get_context_data(**kwargs)
        context['page_title'] = self.PAGE_TITLE
        context['ticket_status_toggle'] = self.TICKET_STATUS_TOGGLE
        context['ticket_status_toggle_url'] = self.TICKET_STATUS_TOGGLE_URL
        context['display_dev_filter'] = self.DISPLAY_DEV_FILTER
        context['display_status_filter'] = self.DISPLAY_STATUS_FILTER
        return context

    def get_queryset(self):
        return models.Ticket.objects.filter_for_team(self.kwargs['team_pk']).filter_for_user(self.request.user).exclude(status='closed')

Edit 3:编辑3:

I found a solution to access the URL kwargs without modifying context data.我找到了一种无需修改上下文数据即可访问 URL kwargs 的解决方案。 I'm not sure why team_pk=team_pk didn't work in the URL tag, but team_pk=view.kwargs.team_pk does work.我不确定为什么team_pk=team_pk在 URL 标签中不起作用,但team_pk=view.kwargs.team_pk确实起作用。

Based on the comments and responses from Willem Van Onsem, a friend of mine, and some of my own digging, I found what I think is the DRYest way to do what I was trying to do.根据我的朋友 Willem Van Onsem 的评论和回复,以及我自己的一些挖掘,我发现了我认为最干的方法来做我想做的事情。 I created a custom mixin:我创建了一个自定义 mixin:

class CommonTemplateContextMixin:
    def get_context_data(self, **kwargs):
        context = super().get_context_data(**kwargs)
        if self.kwargs.get('team_pk'):
            context.setdefault('team_pk', self.kwargs.get('team_pk'))
        else:
            context.setdefault('team_pk', self.request.user.teams.all()[0].pk)
        return context

Then I subclass this mixin with all my views.然后我用我所有的观点对这个 mixin 进行子类化。 Then I have a team_pk template tag inside all my templates, and I just add team_pk=team_pk to all my URL template tags.然后我的所有模板中都有一个team_pk模板标签,我只需将team_pk=team_pk添加到我所有的 URL 模板标签中。 The only advantage to using this instead of team_pk=view.kwargs.team_pk is that I can add some additional logic for the case when the URL kwarg isn't available.使用它而不是team_pk=view.kwargs.team_pk的唯一优点是,当 URL kwarg 不可用时,我可以添加一些额外的逻辑。

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