Python Flask，TypeError:“dict"对象不可调用 [英] Python Flask, TypeError: 'dict' object is not callable

问题描述

有一个似乎很常见的问题，但我已经完成了我的研究，并没有看到它在任何地方完全重现.当我打印 json.loads(rety.text) 时，我看到了我需要的输出.然而，当我调用返回时，它向我显示了这个错误.有任何想法吗?非常感谢并感谢您的帮助.我正在使用 Flask MethodHandler.

Having an issue that seems to be common yet I have done my research and don't see it being exactly recreated anywhere. When I print json.loads(rety.text), I am seeing the output I need. Yet when I call return, it shows me this error. Any ideas? Help is greatly appreciated and thank you. I am using the Flask MethodHandler.

class MHandler(MethodView):
    def get(self):
        handle = ''
        tweetnum = 100

        consumer_token = '' 
        consumer_secret = ''
        access_token = '-'
        access_secret = ''

        auth = tweepy.OAuthHandler(consumer_token,consumer_secret)
        auth.set_access_token(access_token,access_secret)

        api  = tweepy.API(auth)

        statuses = api.user_timeline(screen_name=handle,
                          count= tweetnum,
                          include_rts=False)

        pi_content_items_array = map(convert_status_to_pi_content_item, statuses)
        pi_content_items = { 'contentItems' : pi_content_items_array }

        saveFile = open("static/public/text/en.txt",'a') 
        for s in pi_content_items_array: 
            stat = s['content'].encode('utf-8')
            print stat

            trat = ''.join(i for i in stat if ord(i)<128)
            print trat
            saveFile.write(trat.encode('utf-8')+'
'+'
')

        try:
            contentFile = open("static/public/text/en.txt", "r")
            fr = contentFile.read()
        except Exception as e:
            print "ERROR: couldn't read text file: %s" % e
        finally:
            contentFile.close()
        return lookup.get_template("newin.html").render(content=fr) 

    def post(self):
        try:
            contentFile = open("static/public/text/en.txt", "r")
            fd = contentFile.read()
        except Exception as e:
            print "ERROR: couldn't read text file: %s" % e
        finally:
                contentFile.close()
        rety = requests.post('https://gateway.watsonplatform.net/personality-insights/api/v2/profile', 
                auth=('---', ''),
                headers = {"content-type": "text/plain"},
                data=fd
            )

        print json.loads(rety.text)
        return json.loads(rety.text)


    user_view = MHandler.as_view('user_api')
    app.add_url_rule('/results2', view_func=user_view, methods=['GET',])
    app.add_url_rule('/results2', view_func=user_view, methods=['POST',])

这是回溯(记住上面打印的结果):

Here is the Traceback(Keep in mind results are printing above):

Traceback (most recent call last):
  File "/Users/RZB/anaconda/lib/python2.7/site-packages/flask/app.py", line 1836, in __call__
    return self.wsgi_app(environ, start_response)
  File "/Users/RZB/anaconda/lib/python2.7/site-packages/flask/app.py", line 1820, in wsgi_app
    response = self.make_response(self.handle_exception(e))
  File "/Users/RZB/anaconda/lib/python2.7/site-packages/flask/app.py", line 1403, in handle_exception
    reraise(exc_type, exc_value, tb)
  File "/Users/RZB/anaconda/lib/python2.7/site-packages/flask/app.py", line 1817, in wsgi_app
    response = self.full_dispatch_request()
  File "/Users/RZB/anaconda/lib/python2.7/site-packages/flask/app.py", line 1478, in full_dispatch_request
    response = self.make_response(rv)
  File "/Users/RZB/anaconda/lib/python2.7/site-packages/flask/app.py", line 1577, in make_response
    rv = self.response_class.force_type(rv, request.environ)
  File "/Users/RZB/anaconda/lib/python2.7/site-packages/werkzeug/wrappers.py", line 841, in force_type
    response = BaseResponse(*_run_wsgi_app(response, environ))
  File "/Users/RZB/anaconda/lib/python2.7/site-packages/werkzeug/test.py", line 867, in run_wsgi_app
    app_rv = app(environ, start_response)

推荐答案

Flask 只期望视图返回一个类似响应的对象.这意味着一个 Response、一个字符串或一个描述主体、代码和标题的元组.您正在返回一个 dict，这不是其中之一.由于您要返回 JSON，因此请返回正文中包含 JSON 字符串且内容类型为 application/json 的响应.

Flask only expects views to return a response-like object. This means a Response, a string, or a tuple describing the body, code, and headers. You are returning a dict, which is not one of those things. Since you're returning JSON, return a response with the JSON string in the body and a content type of application/json.

return app.response_class(rety.content, content_type='application/json')

<小时>

在您的示例中，您已经有一个 JSON 字符串，即您发出的请求返回的内容.但是，如果要将 Python 结构转换为 JSON 响应，请使用 jsonify:

data = {'name': 'davidism'}
return jsonify(data)

<小时>

在幕后，Flask 是一个 WSGI 应用程序，它期望传递可调用对象，这就是为什么您会收到特定错误的原因:dict 不可调用且 Flask 不知道如何将其转换为可调用对象.

---

Behind the scenes, Flask is a WSGI application, which expects to pass around callable objects, which is why you get that specific error: a dict isn't callable and Flask doesn't know how to turn it into something that is.

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