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。
json.loads(rety.text)
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)
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。
Response
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。