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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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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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()并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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