小编典典

Python Flask,TypeError:“ dict”对象不可调用

flask

有一个似乎很常见的问题,但我已经完成了研究,并且看不到它在任何地方都被完全重新创建了。当我打印时json.loads(rety.text),我看到了所需的输出。但是,当我打电话给return时,它显示了这个错误。有任何想法吗?非常感谢你的帮助,谢谢。我正在使用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')+'\n'+'\n')

        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Traceback(请记住上面的结果在打印):

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)

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2020-04-05

共1个答案

小编典典

Flask只希望视图返回类似响应的对象。 这表示Response描述主体,代码和标题的,字符串或元组。你正在返回一个dict,这不是其中之一。由于你要返回JSON,因此请返回响应,该响应的正文中包含JSON字符串,内容类型为application/json

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

在你的示例中,你已经有一个JSON字符串,即你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)

在幕后,Flask是一个WSGI应用程序,它期望传递可调用对象,这就是为什么你会收到该特定错误的原因:dict是不可调用的,并且Flask不知道如何将其转换为可调用对象。

2020-04-05