I am doing post request from vue and trying to collect the data in Django views But the part that receiving data in views isn't quite working.
here is views.py
from ..models import project
def insertProjects(request):
projectModel = project.Project()
if request.method == "POST":
#projectModel.project_id = request.POST.get("project_id")
#projectModel.project_name = request.POST.get("project_name")
#projectModel.company_name = request.POST.get("company_name")
#projectModel.save()
name = request.POST.get("project_name")
print(name)
return HttpResponse(True)
After vue POST request called. The print() is returns None? But request payload is not empty. And endpoint returns 200
Any idea about this problem? How can I solve this issue.
If your issue is because you are passing many elements in each key of your request and the view.py is handling only the last element of each key, you should try to use getlist:
project_name_list = request.POST.getlist("project_name")
for name in project_name_list:
projectModel.project_name = name
Related
I have the following python code in my Django views.py, the code takes in a JSON body and send the extracted DATA to another API endpoint, I have simplified the code here.
How do I enable csrf such that it will send the token back to the caller for this method? I am calling this from postman.
#csrf_protect
def validate_booking(request):
if request.method != "POST":
return HttpResponseServerError("Invalid HTTP method")
body = json.loads(request.body)
booking_details = body["booking_details"]
DATA = {
"name": booking_details["name"],
"nric": booking_details["nric"],
"booking_id": booking_details["booking_id"]
}
return HttpResponse(status="200")
This site directs to put this piece of code in my method. But what is "a_template.html"?
https://docs.djangoproject.com/en/4.1/ref/csrf/
#csrf_protect
def my_view(request):
c = {}
# ...
return render(request, "a_template.html", c)
This isn't an easy thing to do as CSRF is 2 steps thing
There is a value that is passed to the client and it is saved to the session on the server.
When a POST request is received, the client shall send this as csrfmiddlewaretoken in the body and the server will check the value against the stored one in the server's session.
So this isn't feasible to be done in APIs as you require session Management which is not of REST API implementations.
Thanks for your reply. I managed to find a solution by doing the following:
Create a new GET method that will generate the session CSRF token using python
Instead of using render which expects a HTML template file, I used JsonResponse(data) to return in JSON format directly
In my postman app which I am making the POST request with the X-CSRFToken in the header, I will first make a GET request to the new method I created in step 1 to retrieve the token and store it as an environment variable
The following is the GET method sample:
from django.http import JsonResponse
def get_csrf_token(request):
csrf_token = csrf(request)['csrf_token']
data = {'csrf_token': csrf_token}
return JsonResponse(data)
I have urls that take a parameter called board_slug. Before getting the template the view will replace the slug if its name is wrong and redirect it. I have already made the code for fixing the slug but do not know how insert the fixed board_slug into the new url. This code is run across multiple views so it has to work with all the following kinds urls:
url(r'^boards/(?P<board_slug>[^/]+)/$', views.BoardView.as_view(), name='board'),
url(r'^/front-thing/boards/(?P<board_slug>[^/]+)/new/$', views.BoardView.as_view(), name='new_board'),
url(r'^boards/(?P<board_slug>[^/]+)/etc...$', views.BoardView.as_view(), name='anything_with_board'),
class BoardView(View):
template_name = 'forums/board.html'
def get(self, request, board_slug):
if wrong:
url = get_django_url
url.board_slug = 'new-slug'
return redirect(url)
else:
return template
I have created a api which i am sending request by post but didn't get vairable in a view
def login_list(request):
if request.method == 'POST':
data = json.dumps(request.POST)
print(data)
serializer = LoginSerializer(data=request.data)
#print(serializer)
return JsonResponse({"message":'fdsafdsa'})
when i print data print(data) then out put is coming like this
{"{\"login\":122122,\"abvc\":\"544545\"}": ""}
and i calling this api like this in postman
Post http://localhost:8000/login/login/
{"login":122122,"abvc":"544545"}
I am not geting value with this
print(request.POST['login']);
how can i get value
Try request.data instead of request.POST. JSON Content is sent in body, which is parsed by Django Rest Framework at runtime.
login_variable = request.data['login']
And ensure you have added 'JSONParser' in REST_FRAMEWORK settings.
In the twilio example made in flask, I can send an SMS and receive an answer using text and the SMS as the search parameter in the database. I need make this but in a django project, my first option that I thought make a django view with a parameter using a url with for send the parameter, but I see that is bad idea because not is possible can use the text of SMS as parameter
This is a part of flask example
#app.route('/directory/search', methods=['POST'])
def search():
query = request.form['Body']
I need make some similar to that view in django using django restframework but how I can get the Body (I think that the body is the text send in the SMS)
for use this as parameter
Use request.POST to access the form data:
from django.shortcuts import render
def my_view(request):
if request.method == "POST":
data = request.POST
# all posted data
data['body']
# the rest of your view logic
return render(request, 'template.html', context)
My views.py code:
from django.template import Context, loader, RequestContext
from django.http import HttpResponse
from skey import find_root_tags, count, sorting_list
from search.models import Keywords
def front_page(request):
if request.method == 'get' :
str1 = request.getvalue['word']
fo = open("xml.txt","r")
for i in range(count.__len__()):
file = fo.readline()
file = file.rstrip('\n')
find_root_tags(file,str1,i)
list.append((file,count[i]))
sorting_list(list)
for name, count in list:
s = Keywords(file_name=name,frequency_count=count)
s.save()
fo.close()
return HttpResponseRedirect('/results/')
else :
str1 = ''
list = []
template = loader.get_template('search/front_page.html')
c = RequestContext(request)
response = template.render(c)
return HttpResponse(response)
def results(request):
list1 = Keywords.objects.all()
t = loader.get_template('search/results.html')
c = Context({'list1':list1,
})
return HttpResponse(t.render(c))
#this for everyone.
the flow is this:
1) I run my app on the server .
2)It shows me the search page due to the else part of the view "def front_page(request)", now I want to execute the if part of the view "def front_page(request)" because I want to execute my python code written there and the redirected to the view "def results(request)", how can I do that ?
3) what should I mention in "action" of the front_page.html and in urls.py so that I can get back to the same view again. because I could'nt get back to the same view that I want it is repetitively showing me the same search page.Please help.
To enlarge upon the answer posted by #Barnaby....by using action='#' your form will be posted to the same url as the url used in the get request for the form.
Then in your view code, you have logic that says - if the request for this url is a GET request then do the work to configure the form, otherwise, you assume it is a POST and then you can handle the response.
Additionally I would advise that the your view explicitly checks that the request is a POST and if not make the assumption that it is a GET, rather than the other way around (as you have it), this is safer, as GET and POST are not the only request types, and you definitely need to know that you are dealing with a POST request if you want to deal with variables submitted in the POST request.
Hope that helps
Short answer: action="#". This is a HTML trick to post back to the current URL.
The general answer to how to reference a view in a template is to use the url tag. You may also want to consider using Django's forms functionality.