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Python Flask,TypeError:“dict”对象不可调用

[英]Python Flask, TypeError: 'dict' object is not callable

Having an issue that seems to be common yet I have done my research and don't see it being exactly recreated anywhere.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。 When I print json.loads(rety.text) , I am seeing the output I need.当我打印json.loads(rety.text) ,我看到了我需要的输出。 Yet when I call return, it shows me this error.然而,当我调用返回时,它向我显示了这个错误。 Any ideas?有任何想法吗? Help is greatly appreciated and thank you.非常感谢并感谢您的帮助。 I am using the Flask MethodHandler .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',])

Here is the Traceback(Keep in mind results are printing above):这是回溯(请记住上面打印的结果):

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 only expects views to return a response-like object. Flask 只期望视图返回一个类似响应的对象。 This means a Response , a string, or a tuple describing the body, code, and headers.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。 You are returning a dict, which is not one of those things.您正在返回一个 dict,这不是其中之一。 Since you're returning JSON, return a response with the JSON string in the body and a content type of application/json .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。

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

In your example, you already have a JSON string, the content returned by the request you made.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。 However, if you want to convert a Python structure to a JSON response, use jsonify :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)

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.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:dict 不可调用且 Flask 不知道如何将其转换为可调用对象。

Use the Flask.jsonify function to return the data.使用 Flask.jsonify 函数返回数据。

from flask import jsonify 
# ...
return jsonify(data)

If you return a data, status, headers tuple from a Flask view, Flask currently ignores the status code and content_type header when the data is already a response object, such as what jsonify returns.如果从 Flask 视图返回data, status, headers元组,当数据已经是响应对象(例如jsonify返回的内容)时,Flask 当前会忽略状态代码和content_type标题。

This doesn't set the content-type header:这不会设置内容类型标头:

headers = {
    "Content-Type": "application/octet-stream",
    "Content-Disposition": "attachment; filename=foobar.json"
}
return jsonify({"foo": "bar"}), 200, headers

Instead, use flask.json.dumps to generate the data (which is what jsonfiy uses internally).相反,使用flask.json.dumps生成数据(这是jsonfiy内部使用的)。

from flask import json

headers = {
    "Content-Type": "application/octet-stream",
    "Content-Disposition": "attachment; filename=foobar.json"
}
return json.dumps({"foo": "bar"}), 200, headers

Or work with the response object:或者使用响应对象:

response = jsonify({"foo": "bar"})
response.headers.set("Content-Type", "application/octet-stream")
return response

However, if you want to literally do what these examples show and serve JSON data as a download, use send_file instead.但是,如果您真的想按照这些示例显示的内容进行操作并将 JSON 数据作为下载提供,请改用send_file

from io import BytesIO
from flask import json
data = BytesIO(json.dumps(data))
return send_file(data, mimetype="application/json", as_attachment=True, attachment_filename="data.json")

as for flask version 1.1.0 now you could return dict至于烧瓶版本 1.1.0 现在你可以返回 dict

flask will convert it automatically to json response. flask 会自动将其转换为 json 响应。

https://flask.palletsprojects.com/en/1.1.x/quickstart/#apis-with-json https://flask.palletsprojects.com/en/1.1.x/changelog/#version-1-1-0 https://flask.palletsprojects.com/en/1.1.x/quickstart/#apis-with-json https://flask.palletsprojects.com/en/1.1.x/changelog/#version-1-1-0

这没有尝试对响应进行 jsonify,而是有效。

return response.content

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