如何在 ModelResource django-import-export 中获取请求

kod*_*ory 4 python django python-3.x django-import-export

如何从 django-import-export 中 ModelResource 的方法查询集中获取 request.user?

class PeopleResource(ModelResource):
    class Meta:
       model = People
       exclude = ('id','agent', 'public_id', 'active')

    def dehydrate_placeA(self, people):
        ...
        ...

    def get_queryset(self):
        query = People.objects.filter( ..... request.user )
        return query
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tim*_*haw 5

我正在寻找类似的东西。我希望请求对象(基于在 admin 中发出的要导出的请求)附加到 Resource 实例,以便我可以检查它并根据查询参数动态影响功能。如果您想根据用户动态更改它,这也将非常有用。结果很简单:

首先,继承 ModelResource 类并寻找新的 kwarg:

from import_export import resources

class RequestModelResource(resources.ModelResource):
    def __init__(self, *args, **kwargs):
        self.request = kwargs.pop('request', None)
        super(RequestModelResource, self).__init__(*args, **kwargs)
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然后,在 import-export 中有一个相关的管理方法,您可以使用它来传递 kwargs。见这里。将此添加到您的 ModelAdmin 继承自import_export.admin.ImportExportModelAdmin

class MyModelAdmin(ImportExportModelAdmin):
    resource_class = MyModelResource

    def get_resource_kwargs(self, request, *args, **kwargs):
        """ Passing request to resource obj to control exported fields dynamically """
        return {'request': request}
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基本上就是这样。现在可以在从 RequestModelResource 继承的 Resource 类中的任何位置使用请求。例如:

class MyModelResource(RequestModelResource):
    def get_export_fields(self):
        fields = super().get_fields()

        # Check self.request.user, self.request.GET, etc to impact logic
        # however you want!

        return fields
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kod*_*ory 2

问题解决了。我忘记我people_resource.export()之前已经表达过我的观点了。现在更容易了,因为我只需在 method 中传递 variabel kwargsexport()def export()在 model 中重写PeopleResource。所以我的解决方案就在这里。

视图.py

def export_excel(request):    
    ...
    people_resource = PeopleResource()
    dataset = people_resource.export(agent=request.user.agent,)
    ...
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模型.py

class PeopleResource(ModelResource):
    class Meta:
        model = People
        exclude = ('id','agent', 'public_id', 'active')
    ...
    ...

    def export(self, queryset=None, *args, **kwargs):
        queryset = People.objects.filter(agent=kwargs['agent'])
        return super(PeopleResource, self).export(queryset, *args, **kwargs)
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