url_for 对于 Flask-Admin 中基于 class 的视图

Question

我有一个基于 class 的管理视图：

class All_RDPs(BaseView):
    @expose('/')
    def index(self):
        return 'ok1'
    @expose('/test')
    def testindex(self):
        return 'ok2'

像这样在 Flask-Admin 中注册的：

admin.add_view(All_RDPs(name='dep_rdp'))

然后可以像这样从浏览器中查看：

http://localhost/admin/all_rdps/
http://localhost/admin/all_rdps/test

问题是：

如何为此 class 指定 URL 而不是默认生成的名称 all_rdps？
如何使用 url_for 为这些端点生成 url？ url_for('admin.All_RDPs.testindex')、url_for('admin.All_RDPs') 不工作。

Answer 1

You can override the endpoint name by passing endpoint parameter to the view class constructor:
admin = Admin(app)
admin.add_view(MyView(endpoint='testadmin'))
In this case, you can generate links by concatenating the view method name with an endpoint:
url_for('testadmin.index')
If you don't override the endpoint name, the lower-case class name can be used for generating URLs, like in:
url_for('myview.index')
For model-based views the rules differ - the model class name should be used if an endpoint name is not provided. The ModelView also has these endpoints by default: .index_view, .create_view, and .edit_view. So, the following urls can be generated for a model named "User":
# List View
url_for('user.index_view')

# Create View (redirect back to index_view)
url_for('user.create_view', url=url_for('user.index_view'))

# Edit View for record #1 (redirect back to index_view)
url_for('user.edit_view', id=1, url=url_for('user.index_view'))

来源：Flask-Admin quickstart

url_for 对于 Flask-Admin 中基于 class 的视图

url_for for class-based views in Flask-Admin

python

flask

flask-admin