如何在每个函数调用中使用带有Flask-RESTPlus的装饰器？

Question

我有问题将装饰器应用到我的Resource.该Api decorators参数被应用到每个资源功能或怎么做我已经明白了decorators参数？我的装饰器仅在启动时应用,而不是在每个函数调用时应用.我究竟做错了什么？

from flask import Flask, Blueprint
from flask.ext.restplus import Api, Resource, fields

app = Flask(__name__)

def my_decorator(input):
    print("my decorator")
    return input

api_blueprint = Blueprint('index', __name__, url_prefix='/0')

api = Api(api_blueprint, version='0.1.0', title='Index API',
    description='The Index API helps to find things faster in the own database',
    # ui=False,
    decorators=[my_decorator]
)

app.register_blueprint(api_blueprint)

ns = api.namespace('index', description='Index API')


search_id_score_model = api.model('id', {
    'id': fields.String(required=True),
    'score': fields.Float(required=True)
})

search_result_model = api.model('result', {
    'key': fields.String(required=True),
    'ids': fields.List(fields.Nested(search_id_score_model, required=True), required=True)
})


@ns.route('/search/<string:query>')
class Search(Resource):
    '''Serach'''

    @api.marshal_list_with(search_result_model)
    @my_decorator
    def get(self, query):
        '''fetch result by query'''
        return None

if __name__ == '__main__':
    app.run(debug=True)

Answer 1

有两个装饰器级别:API范围和资源范围.

Api.decorators参数允许将装饰器应用于所有资源,而Resource.method_decorators将应用于给定资源的所有方法.

在您的情况下,您的装饰器只在应用后调用打印而不是一次调用的问题.你应该像这样修改装饰器:

def my_decorator(func):
    def wrapper(*args, **kwargs):
        print("my decorator")
        return func(*args, **kwargs)
    wrapper.__doc__ = func.__doc__
    wrapper.__name__ = func.__name__
    return wrapper

注意__doc__和__name__影响.第一个允许正确提取描述,__name__允许避免自动命名冲突.如果为每个资源手动指定端点,则该名称是可选的.