Django - 将模型名称从 url 传递给视图

Posted 2023-02-24

【中文标题】Django - 将模型名称从 url 传递给视图

【英文标题】：Django - Passing the model name to a view from url

【发布时间】：2014-07-03 13:03:17

【问题描述】：

我将如何做到这一点，以便我可以将模型名称作为 url 中的参数传递给视图？我想重用这个视图，只传递模型名称以显示参数是什么模型的列表。

这就是我目前所拥有的

查看

class ModelListView(ListView,objects):

    model = objects
    template_name = "model_list.html"

    def get_context_data(self,**kwargs):
        context = super(ModelListView, self).get_context_data(**kwargs)
        context['listobjects'] = model.objects.all()
        return context

网址

url(r'^musicpack', MusicPackListView.as_view(), name='musicpack-list', objects = 'MusicPack'),
url(r'^instruments', MusicPackListView.as_view(), name='instrument-list', objects = 'Instrument'),

已回答

谢谢你的回答

我已经完成了以下操作，它似乎有效。

查看

class ModelListView(ListView):

    template_name = "model_list.html"

    def get_context_data(self,**kwargs):
        context = super(ModelListView, self).get_context_data(**kwargs)
        return context

网址

#models
from inventory.views import MusicPack
from inventory.views import Instrument
#views
from inventory.views import ModelListView

    url(r'^musicpacks', ModelListView.as_view(model = MusicPack,), name='musicpack-list'),
    url(r'^instruments', ModelListView.as_view(model = Instrument,), name='instrument-list'),

【参考方案1】：

我会将参数从 url 传递到如下视图：

观看次数：

class ModelListView(ListView):

    model = None
    model_name= ''
    object = None

    template_name = "model_list.html" 

    def get_context_data(self,**kwargs):
        context = super(ModelListView, self).get_context_data(**kwargs)
        context['listobjects'] = model.objects.all()
        return context

网址：

url(r'^musicpack', ModelListView.as_view( model= MusicPackList,model_name= 'music_pack_list' object = 'MusicPack')),
url(r'^instruments', ModelListView.as_view( model=InstrumentPackList,model_name= 'instrument_pack_list', object= 'InstrumentPack'))