I’m currently building a little Django Project and I’ve got a question. I would like to make an API Call to an external API with a user input. How can I achieve that? Pretty new to Django sorry.
Regards,
You can make an API view in Django to receive the request from the user whos using your system with parameters from request body or header (user input) using API View following this link: Django API View.
You can implicitly take these params from the user and send them to the other system to make an API request using requests in python
for example: Pyhton requests
Related
I want to call an external service API in the Django admin and display the data.
Also, I want to create a form so that taking input from the admin, should able to make a POST to external service.
My app does not have any models.
django-admin version 2.2.8
Please, someone, provide a solution stuck from last few days
Thank You.
For the first question, regarding the external service, you should use the python requests package. Just pip install requests.
For the second one, when you have a form inside the admin, you can access the form's data using request.POST and then use the aforementioned requests package to do a server-side POST request to your external API.
Lastly, you don't need models. An admin GUI is of its basic nature. In Django, you can replace the admin template for your own by specifying where your template lives. You can follow along with this: How to override and extend basic Django admin templates?
I want to build a project with Django framework, it includes website and mobile app, website is built based on Django template, and mobile app needs rest api. How can I do it with only one copy of source code?
In other words, can I create one project, in which there are several django apps, it supports returning both rendering template and json for mobile app?
I know we can use Angular JS in website, then both website and mobile app access rest api build with Django-Rest-Framework. But I have no idea about Angular JS and there is no time for me to learn it.
Please help.
This is easy to achieve if you leverage the power of HTTP headers, specifically speaking the "Content-Type" header
(HTTP 1.1 headers specification)
The way I personally use it is something like this:
def my_view_name(request):
if 'CONTENT_TYPE' in request.META and \
request.META['CONTENT_TYPE'] == 'application/json':
return HttpResponse(
json.dumps({"foo":"bar"}),
content_type='application/json'
)
else:
return render_to_response(
'sometemplate.html', context
)
The code above is django-specific.
It allows you to control what kind of response the server doles out, based out what the client passes in. It can detect that a JSON response is needed by checking the Content-Type header.
Just ensure your mobile app makes its HTTP requests with the Content-Typeheader set to application/json
Hope that's what you wanted.
You can do it in one project. Have the same Model, but have different views for website and REST-api.
You can build one Common Backend to serve your data.
And you can do this without having a Django-rest-Framework(without having to use angular).
You are just going call the url of the backend.
For eg.,. lets say backend code is running at 127.0.0.1/8000, and you need to get a user's details, you are just going to call url 127.0.0.1:8000/get_user_details/ from mobile app or any other front end.
And please make a habit of forming/arranging data written in a different function so that it is reusable by both the Mobile app and the website.(plz see below)
### For browser website written in DJANGO (served in template by django):
def user_details_page(request):
get_user_details(someid) #### returns json of user details
render in template
#### And for the mobile App you could use **get_user_details(someid)** just to get json data.
def get_user_details(someid):
###do watever
return json ## of user details
And you can also use Django Rest Framework(without having to use angular), it gives you a lot of stability.
Or you can setup a entire new backend in Django Framework to serve data for mobile App.
I have one django app and few small Flask webservices.
In Flask apps I need to validate if the client logged in Django app and grab his pk if possible.
It seems to be possible by taking session ID from a cookie and manually looking into session storage, but I am looking for some less low-level solution.
You need to build a REST api to do that. In the api on the django side you would query the user by whatever criteria you provided and return user.is_authenticated() and user.pk. Take a look at django-rest-framework or tastypie. Then on Flask app you just hit the api and you are done.
I am writing a Django management command that visits a few pages, logged in as a superuser, and saves the results to a set of .html files.
Right now I'm just using the requests library and running the command with the development server running. Is there an easy way to generate the HTML from a view response so I do without actual HTTP requests?
I could create a request object from scratch but that seems like more overhead than the current solution. I was hoping for something simple.
Django has a RequestFactory which seems to suit your needs.
While it's not exactly meant for this purpose, an option would be to use the testing framework's Client to fake a request to the url - be sure to use client.login() before making your requests, to ensure you have superuser capabilities.
I am currently building a simple django app to serve basic models to an Android client. I'm using Tastypie as webservice, and I would like (well.. I need to) to implement a minimal security process as follow :
Hardcode the secret key in the Android app. Done.
Request via http GET the data I want, with the secret key as parameter. Done
Make the tastypie check this key in the requests made to django by the android app. ?..
I guess it is a fairly easy thing to do but I'm currently learning Django / Python etc and I haven't figured out what to make of the tastypie docs on this particular topic :
http://django-tastypie.readthedocs.org/en/latest/authentication_authorization.html
FYI, I'd rather not have this key user-generated for the simple reason I did not understand how it would work. Tastypie says it can generate an api key upon user creation, but what then ? How does the app get this key ?
Thank you very much for your help !
I suggest you look at django middleware. I'm assuming you're passing the key through in the request (I'd do it in the request header). The middleware should simply check for the existance of this key, and if it exists, let the request through, otherwise return a 403.