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を失敗しましたStacktrace:Djangoは管理者list_displayに外部キーを逆に私は、管理list_change上の逆の外部キーを表示するには、トリングだ
File "/venv/lib/python2.7/site-packages/django/contrib/admin/templatetags/admin_list.py", line 320, in result_list
'results': list(results(cl))}
File "/venv/lib/python2.7/site-packages/django/contrib/admin/templatetags/admin_list.py", line 296, in results
yield ResultList(None, items_for_result(cl, res, None))
File "/venv/lib/python2.7/site-packages/django/contrib/admin/templatetags/admin_list.py", line 287, in __init__
super(ResultList, self).__init__(*items)
File "/venv/lib/python2.7/site-packages/django/contrib/admin/templatetags/admin_list.py", line 199, in items_for_result
f, attr, value = lookup_field(field_name, result, cl.model_admin)
File "/venv/lib/python2.7/site-packages/django/contrib/admin/utils.py", line 278, in lookup_field
value = attr(obj)
File "/home/kparsa/boom/myapp/myapp/admin.py", line 299, in links
link = obj.book_set().all()
File "/venv/lib/python2.7/site-packages/django/db/models/fields/related.py", line 691, in __call__
manager = getattr(self.model, kwargs.pop('manager'))
KeyError: u'manager'
これを正しく動作させる方法を知っている人はいますか?私はそれが多くの重複クエリ(行の#を)実行するため
qres = Book.objects.filter(person__id=obj.id).values_list('id', 'name').order_by('name')
for x, y in qres:
links.append('<a href="/admin/mypp/book?q=ID{}">{}</a>'\
.format(x, y))
ような何かをしたくない 注意。