I have the url and corresponding view as follows.
url(r'^(?P<token>.*?)/ack$', views.api_ACK, name='device-api_ack')
def api_ACK(request, token):
"""
Process the ACK request comming from the device
"""
logger.info('-> api_ACK', extra={'request': request, 'token' : token, 'url': request.get_full_path()})
logger.debug(request)
if request.method == 'GET':
# verify the request
action, err_msg = api_verify_request(token=token, action_code=Action.AC_ACKNOWLEDGE)
return api_send_answer(action, err_msg)
I want to call api_ACK function with request method as GET from another view api_send_answer
I am creating one url in /device/LEAB86JFOZ6R7W4F69CBIMVBYB9SFZVC/ack in api_send_answer view as follows..
def api_send_answer(action, err_msg, provisional_answer=None):
last_action = create_action(session,action=Action.AC_ACKNOWLEDGE,token=last_action.next_token,timer=500)
url = ''.join (['/device/',last_action.next_token ,'/',Action.AC_ACKNOWLEDGE])
logger.debug('Request Url')
logger.debug(url)
response = api_ACK(request=url,token=last_action.next_token) # This is wrong
Now from api_send_answer it is redirecting to api_ACK view, but how to call api_ACK with request method as GET?
Please help..Any suggestions would be helpful to me
This line
response = api_ACK(request=url,token=last_action.next_token) is wrong because view expects HttpRequest object and you give him url instead.
if you need to return view response to user, you can use redirect:
def api_send_answer(action, err_msg, provisional_answer=None):
last_action = create_action(session,action=Action.AC_ACKNOWLEDGE,token=last_action.next_token,timer=500)
url = ''.join (['/device/',last_action.next_token ,'/',Action.AC_ACKNOWLEDGE])
logger.debug('Request Url')
logger.debug(url)
return HttpResponseRedirect(url)
if you need to do something else with view response you have to use HttpRequest object not url as parameter.
Related
This is how my login method looks like:
def login():
if request.method == "GET":
return render_template('user/login.html')
else:
jwt_token = "xxxx"
resp = redirect(url_for('home.index'))
resp.headers.set('Authorization', "JWT {}".format(jwt_token))
return resp
This works fine but the Authorization header does not make it to home.index page.
How do I tell flask to preserve this header on every request?
------Edit---------
This works if I add Token to Cookie as resp.set_cookie('Authorization', "JWT {}".format(json_data["access_token"])) but I would like to keep it in the Authorization Header.
If you want to set Authorization header on all requests, you could do something like this:
#app.after_request
def after_request(response):
my_jwt_token = 'xxxx'
response.headers['Authorization'] = my_jwt_token
return response
More information on documentation: http://flask.pocoo.org/docs/1.0/api/#flask.Flask.after_request
I'm using Django 2.0
I am making Ajax request to FormView to render template inside <div> and want to redirect in certain case.
<script>
function loadNextQuestion() {
$('#question-box').load("{% url 'learn:question' course_learn.pk %}?session="+$('#session-id').val(), function(){
// other stuffs
});
}
</script>
and the view is like
class LearnQuestion(FormView):
form_class = SessionForm
template_name = 'learn/learn_question.html'
def get_context_data(self, **kwargs):
context = super(LearnQuestion, self).get_context_data(**kwargs)
course_learn = CourseLearn.objects.get(pk=self.kwargs['course_learn_id'])
session = self.request.GET['session']
question, question_type, options, complete = CourseLearn.objects.get_next_question(course_learn, session)
if complete:
context['complete'] = complete
context['course_learn'] = course_learn
context['session'] = session
return context
context['question'] = question
context['question_type'] = question_type
context['options'] = options
context['course_learn'] = course_learn
context['session'] = session
context['complete'] = complete
return context
def render_to_response(self, context, **response_kwargs):
if context['complete']:
return redirect(reverse('learn:success',
kwargs={
'course_learn_id': context['course_learn'].pk,
'session': context['session']
}))
return super(LearnQuestion, self).render_to_response(context, **response_kwargs)
which renders learn/learn_question.html template inside question-box <div> but redirects when context['complete'] is True
This works good but on redirect, the template of redirected URL is rendering in same <div id="question=box">
How can I redirect to a complete new page instead of rendering in same <div>?
The redirection won't work. Since normally when you return the HTTP response Redirect, a special kind of HTTP response is sent to browser. Which it interprets and redirects.
This works only when making a traditional request since, in that case, browser handles the response.
Here the situation is that, you are making an AJAX call.
So you have to check the response code of what you receive in JavaScript.
https://stackoverflow.com/a/17436256/4929982
$.ajax( url [, settings ] )
.always(function (jqXHR) {
console.log(jqXHR.status);
});
Then redirect in case it is redirect ( code: 301 or 302).
Redirect using:
window.location.replace('your_url') ;
You should get your_url from the response object.
I am trying to write a small restful api application, i am using Chrome Postman extension for sending requests to the app .
I believe that my code does not have mistakes but every time i am sending post request a 400 Bad Request error raising , here is my code:
#api_route.route('/api', methods=['GET'])
def api():
return jsonify({'message':'Api v1.0'})
#api_route.route('/api', methods=['POST'])
def create_user():
data = request.get_json()
if data:
hashed_password = generate_password_hash(data['password'], method='sha256')
api = Api(email=data['email'], password=hashed_password)
db.session.add(api)
db.session.commit()
return jsonify({'message', 'New User Created!'})
The json data that i am sending looks like this:
{"email" : "Test", "password" : "123123123"}
Why i am getting the 400 error ??
Update:
Screenshots for the requests using Postman:
GET Request
POST Request
Here i am initiating api route inside api controller :
from flask import Blueprint
api_route = Blueprint(
'api',
__name__
)
from . import views
then i am registering it inside def create_app() function :
from .api import api_route
app.register_blueprint(api_route)
Here are the extensions that i am using in my application:
toolbar = DebugToolbarExtension()
assets_env = Environment()
cache = Cache()
moment = Moment()
htmlminify = HTMLMIN()
csrf = CSRFProtect()
jac = JAC()
googlemap = GoogleMaps()
session = Session()
principal = Principal()
I solved the problem, i've initiated CSRFProtect with app so i need to include X-CSRFToken in all my requests, so i have two choices:
1 - To include the csrf_token in request.headers for all the requests
2 - Using #csrf.exempt decorator that coming with flask_wtf.csrf
For now i am using #csrf.exempt, so it become like this:
#api_route.route('/api', methods=['GET','POST'])
#csrf.exempt
def create_user():
if request.method == 'GET':
return jsonify({'message' : 'API v1.0'})
elif request.method == 'POST':
data = request.get_json()
hashed_password = generate_password_hash(data['password'], method='sha256')
new_user_api = Api(email=data['email'], password=hashed_password)
db.session.add(new_user_api)
db.session.commit()
return jsonify({'message' : 'New user created!'})
return return jsonify({'message' : 'No user has been added!'})
Thanks for #MrPyCharm for his interests , salute :) .
A good approach would be to structure your views as follows:
Instead of creating view with same route for different request methods, you can handle the request methods in the same view:
#api_route.route('/api', methods=['GET', 'POST'])
def api():
if request.method == 'GET':
return jsonify({'message':'Api v1.0'})
else:
data = request.get_json(force=True)
if data:
hashed_password = generate_password_hash(data['password'], method='sha256')
api = Api(email=data['email'], password=hashed_password)
db.session.add(api)
db.session.commit()
return jsonify({'message': 'New User Created!'})
# Just in case the if condition didn't satisfy
return None
A note for anyone else experiencing this with PostMan and Flask - you will also hit a HTTP 404 if your URL in PostMan is HTTPS but your Flask app only handles HTTP.
How can I get the request url in Scrapy's parse() function? I have a lot of urls in start_urls and some of them redirect my spider to homepage and as result I have an empty item. So I need something like item['start_url'] = request.url to store these urls. I'm using the BaseSpider.
The 'response' variable that's passed to parse() has the info you want. You shouldn't need to override anything.
eg. (EDITED)
def parse(self, response):
print "URL: " + response.request.url
The request object is accessible from the response object, therefore you can do the following:
def parse(self, response):
item['start_url'] = response.request.url
Instead of storing requested URL's somewhere and also scrapy processed URL's are not in same sequence as provided in start_urls.
By using below,
response.request.meta['redirect_urls']
will give you the list of redirect happened like ['http://requested_url','https://redirected_url','https://final_redirected_url']
To access first URL from above list, you can use
response.request.meta['redirect_urls'][0]
For more, see doc.scrapy.org mentioned as :
RedirectMiddleware
This middleware handles redirection of requests based on response status.
The urls which the request goes through (while being redirected) can be found in the redirect_urls Request.meta key.
Hope this helps you
You need to override BaseSpider's make_requests_from_url(url) function to assign the start_url to the item and then use the Request.meta special keys to pass that item to the parse function
from scrapy.http import Request
# override method
def make_requests_from_url(self, url):
item = MyItem()
# assign url
item['start_url'] = url
request = Request(url, dont_filter=True)
# set the meta['item'] to use the item in the next call back
request.meta['item'] = item
return request
def parse(self, response):
# access and do something with the item in parse
item = response.meta['item']
item['other_url'] = response.url
return item
Hope that helps.
Python 3.5
Scrapy 1.5.0
from scrapy.http import Request
# override method
def start_requests(self):
for url in self.start_urls:
item = {'start_url': url}
request = Request(url, dont_filter=True)
# set the meta['item'] to use the item in the next call back
request.meta['item'] = item
yield request
# use meta variable
def parse(self, response):
url = response.meta['item']['start_url']
In my view function, I'd like to return a json object (data1) and some text/html (form). Is this possible?
Here is part of my views.py:
if request.is_ajax() and request.method == 'POST':
...
if form.is_valid():
answer = form.cleaned_data['answer'] # Answer extracted from form is also a string
a1 = ques1.correct_answer
if a1 == answer:
test1 = question_list.get(id=nextid)
form = AnswerForm(test1)
ques1 = question_list.filter(id=nextid) # Filter next question as <qs>
data1 = serializers.serialize("json",ques1) # Json-ize
# ********EDITED HERE **********
variables1 = Context({
'form' : form,
'q1' : data1,
})
#response = HttpResponse()
#response['data1'] = response.write(data1)
#response['form'] = response.write(form)
if nextid <= qsnlen:
return HttpResponse(variables1, mimetype="application/json")
#return HttpResponse(response)
else:
...
I'd like to send back both the form html and the ques1 json object. How can I do this? Thanks in advance.
Just put both pieces of data in a JSON container, one key with the form data and one with the HTML as a rendered string. In the browser, you can just pull both keys out & do your thing.
In your view:
form_json_data = get_form_json_data()
rendered_html = get_the_html()
return HttpResponse(json.dumps({
"formdata": form_json,
"html": rendered_html}),
content_type="application/json")
In js:
$.post(foo, postdata, function(data){
var formdata = data.formdata
var html = data.html;
$(".html-target").replaceWith(html);
do_whatever(formdata);
})
Use JsonResponse
from django.http import JsonResponse
response_data = {put your data into a dict}
return JsonResponse(response_data, status=201)
To do this with one response; you need to send the JSON as a plain text in the context of your template response (HTML).
If you need to send JSON as as a separate JSON object, with its own mime type, then you need to write two views; one that sends back the JSON as application/json and the other that sends back the form (HTML).
EDIT:
You are not returning JSON objects, but you are turning a dictionary that has two items of two different types. As I explained in the comments, in one request/response cycle; you can only return one response which has a specific mime type that is based on the content and how you want the browser to handle it. Most of the time the content type is 'text/html'.
In your scenario, if you want to return both the HTML (which is your form), and the JSON response (which is a string), you need to return HTML.
If you want to return JSON to Jquery as a JSON object; you need to detect the request type. In your front end (the templates), you will initiate two requests - one from the browser, which will return back the form. The other from jQuery, which will return the appropriate JSON object.
Here is a possible approach to this:
def foo(request):
if request.is_ajax():
ctx = dict()
ctx['hello'] = 'world'
return HttpResponse(json.dumps(ctx),content_type='application/json')
else:
return HttpResponse('hello world')