Python future : How do I get the json from a future object in Tornado?

Question

这是一个post处理程序:

处理程序.py

from imports import logic

@gen.coroutine
def post(self):
    data = self.request.body.decode('utf-8')
    params = json.loads(data)
    model_id= params['model_id']
    logic.begin(model_id)

logic对象是从imports.py导入的，它是从导入的类Logic实例化的

imports.py:

import Models
import Logic

class Persist(object):
    def getModel(self, model_id):
        model = Models.findByModelId(model_id)
        return model


persist = Persist()
logic = Logic(persist)

逻辑.py

class Logic(object):
    def __init__(self, persist):
        self._persist = persist

    def begin(self, model_id):
         model = self._persist.get_model(model_id)
         print ("Model from persist : ")
         print (model)

get_model 方法使用 Models 进行数据库查询并返回 future 对象:

模型.py:

from motorengine.document import Document

class Models(Document):
    name = StringField(required=True)

def findByModelId(model_id):
    return Models.objects.filter(_id=ObjectId(model_id)).find_all()

这会在控制台中打印 future 的对象:

<tornado.concurrent.Future object at 0x7fbb147047b8>

如何将其转换为 json ？

Answer 1

要将 Future 解析为实际值，请在协程中yield它:

@gen.coroutine
def begin(self, model_id):
     model = yield self._persist.get_model(model_id)
     print ("Model from persist : ")
     print (model)

任何调用协程的函数都必须是协程，并且它必须yield协程的返回值才能获取其返回值:

@gen.coroutine
def post(self):
    data = self.request.body.decode('utf-8')
    params = json.loads(data)
    model_id = params['model_id']
    model = yield logic.begin(model_id)
    print(model)

更高级的编码模式不需要遵循这些规则，但首先要遵循这些基本规则。

有关从协程调用协程的更多信息，请参阅 Refactoring Tornado Coroutines .