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我有一个带有 Flask-Admin 的烧瓶应用程序来管理具有模型视图的用户/admin/userview。我希望它在数据库中具有正确组的用户导航到时显示页面，/admin/userview如果不是，则返回纯文本“非管理员”。幸运的是我得到了最后一部分的工作，但不幸的是我似乎无法得到第一部分（如果他们在正确的组中，继续显示页面）。以下是相关代码：

class MyView(ModelView):
    @expose('/', methods = ['GET', 'POST'])
    def index(self):
    ## grab kerberos username (PROD)
        secuser = request.environ.get('REMOTE_USER')

        adminresult = User.query.filter_by(usrlevel='admin').all()
        adminresult = str(adminresult)
        adminresult = adminresult.strip('[]')

        managerresult = User.query.filter_by(usrlevel='manager').all()
        managerresult = str(managerresult)
        managerresult = managerresult.strip('[]')    

        if secuser in adminresult:
            pass         <---------------\
        elif secuser in managerresult:    |- if a user is apart of either, this gives a ValueError
            pass         <---------------/
        else:
            return "NOT ADMIN" <--------- This works!

##ADMIN
admin = Admin(flaskapp, index_view=MyHomeView(), name="test APP")
admin.add_view(MyView(User, db.session, "Edit Users"))

这是我在用户 inadminresult或 in时得到的回溯managerresult：

Traceback (most recent call last):
  File "/usr/local/lib/python2.7/dist-packages/flask/app.py", line 1836, in __call__
    return self.wsgi_app(environ, start_response)
  File "/usr/local/lib/python2.7/dist-packages/flask/app.py", line 1820, in wsgi_app
    response = self.make_response(self.handle_exception(e))
  File "/usr/local/lib/python2.7/dist-packages/flask/app.py", line 1403, in handle_exception
   reraise(exc_type, exc_value, tb)
  File "/usr/local/lib/python2.7/dist-packages/flask/app.py", line 1817, in wsgi_app
    response = self.full_dispatch_request()
  File "/usr/local/lib/python2.7/dist-packages/flask/app.py", line 1478, in full_dispatch_request
    response = self.make_response(rv)
  File "/usr/local/lib/python2.7/dist-packages/flask/app.py", line 1566, in make_response
    raise ValueError('View function did not return a response')
ValueError: View function did not return a response

如果用户在任一组中，如何让ModelView“用户”显示其内容，如果不是，则只返回“非管理员”文本？我想我已经完成了一半，只是继续似乎是一个问题......

谢谢！

4

1 回答 1

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您实际上应该签入is_accessible：

class MyView(ModelView):
    def is_accessible(self):
        # grab kerberos username (PROD)
        secuser = request.environ.get('REMOTE_USER')

        admins_and_managers = User.query
                                  .filter(User.usrlevel.in_('admin', 'manager'))
                                  # Better yet, filter again on `secuser`
                                  # to avoid loading *every* admin and manager
                                  # on *every* request for this resource
                                  # and then use `.first` or `.one`
                                  .all()

        return secuser in admins_and_managers
于 2014-07-15T20:54:27.350 回答