我正在尝试调整它:
http://flask.pocoo.org/docs/views/
进入蓝图本身(基于我看过的其他蓝图)。将 api 注册从应用程序抽象到蓝图初始化中。这是来自 flask 文档的代码，有一些更改。
这似乎可行:
class MyAPI(MethodView):
def __init__(self, name):
self.name = name
bp = Blueprint(name, __name__)
bp_endpoint = '{0}_api'.format(name)
bp_url = '/{0}/'.format(name)
bp_pk = '{0}_tag'.format(name)
self.register_api(bp, bp_endpoint, bp_url, bp_pk, 'string')
self._blueprint = bp
def register_api(self, blueprint, endpoint, url, pk='id', pk_type='int'):
view_func = self.as_view(endpoint)
blueprint.add_url_rule(url, defaults={pk: None},
view_func=view_func, methods=['GET',])
blueprint.add_url_rule(url, view_func=view_func, methods=['POST',])
blueprint.add_url_rule('{0}<{1}:{2}>'.format(url, pk_type, pk), view_func=view_func,
methods=['GET', 'PUT', 'DELETE'])
def get(self, my_tag):
#... with post, put methods etc.
然后在我的应用程序中我可以这样做:
m = MyAPI('my')
app.register_blueprint(m._blueprint)
这似乎有效，注册 url 这样我就可以得到:
Map([<Rule '/my/' (POST, OPTIONS) -> my.my_api>,
<Rule '/my/<my_tag>' (PUT, HEAD, DELETE, OPTIONS, GET) -> my.my_api>,
<Rule '/static/<filename>' (HEAD, OPTIONS, GET) -> static>,
<Rule '/my/' (HEAD, OPTIONS, GET) -> my.my_api>])
但是，现在转到路线时出现错误(我刚刚尝试了 GET):
Traceback (most recent call last):
File "/media/686e26f8-c6d4-4448-8fe4-c19802726dcb/projects/1_current/private/pycleaver/venv/lib/python2.7/site-packages/flask/app.py", line 1518, in __call__
return self.wsgi_app(environ, start_response)
File "/media/686e26f8-c6d4-4448-8fe4-c19802726dcb/projects/1_current/private/pycleaver/venv/lib/python2.7/site-packages/flask/app.py", line 1506, in wsgi_app
response = self.make_response(self.handle_exception(e))
File "/media/686e26f8-c6d4-4448-8fe4-c19802726dcb/projects/1_current/private/pycleaver/venv/lib/python2.7/site-packages/flask/app.py", line 1504, in wsgi_app
response = self.full_dispatch_request()
File "/media/686e26f8-c6d4-4448-8fe4-c19802726dcb/projects/1_current/private/pycleaver/venv/lib/python2.7/site-packages/flask/app.py", line 1264, in full_dispatch_request
rv = self.handle_user_exception(e)
File "/media/686e26f8-c6d4-4448-8fe4-c19802726dcb/projects/1_current/private/pycleaver/venv/lib/python2.7/site-packages/flask/app.py", line 1262, in full_dispatch_request
rv = self.dispatch_request()
File "/media/686e26f8-c6d4-4448-8fe4-c19802726dcb/projects/1_current/private/pycleaver/venv/lib/python2.7/site-packages/flask/app.py", line 1248, in dispatch_request
return self.view_functions[rule.endpoint](**req.view_args)
File "/media/686e26f8-c6d4-4448-8fe4-c19802726dcb/projects/1_current/private/pycleaver/venv/lib/python2.7/site-packages/flask/views.py", line 83, in view
self = view.view_class(*class_args, **class_kwargs)
**TypeError: __init__() takes exactly 2 arguments (1 given)**
aa 这比我认为的 atm 低一两个级别。任何输入感谢我错过的东西。我认为这可能与 register_api 中的 view_func 最初有关。
编辑:
一个答案
class MyAPI(MethodView):
def __init__(self, name):
self.name = name
bp = Blueprint(name, __name__)
self.endpoint = '{0}_api'.format(name)
self.url = '/{0}/'.format(name)
self.pk = '{0}_tag'.format(name)
self._blueprint = bp
self.register_api(self._blueprint, self.endpoint, self.url, self.pk)
def register_api(self, bp, endpoint, url, pk ='id', pk_type='int'):
view_func = self.__class__.as_view(endpoint)
bp.add_url_rule(url, defaults={pk: None},
view_func=view_func, methods=['GET',])
bp.add_url_rule(url, view_func=view_func, methods=['POST',])
bp.add_url_rule('{0}<{1}:{2}>'.format(url, pk_type, pk), view_func=view_func,
methods=['GET', 'PUT', 'DELETE'])
请您参考如下方法:
我认为你可以让你的初始化方法只有一个参数:
def __init__(self):
bp = Blueprint("what?", __name__) # here
bp_endpoint = '{0}_api'.format(name)
bp_url = '/{0}/'.format(name)
bp_pk = '{0}_tag'.format(name)
self.register_api(bp, bp_endpoint, bp_url, bp_pk, 'string')
self._blueprint = bp
或者在 as_view 中提供足够的值而不修改您的初始化方法。
def register_api(self, blueprint, endpoint, url, pk='id', pk_type='int'):
view_func = self.as_view(endpoint, name="what?") # here
# ... omit ...
但在我看来，在方法 View 中创建蓝图并不是一个好主意。蓝图是一个子应用程序，应该由多个 View 共享。
