Django | Triggering a function from list in template - django

I'm trying to make a facebook like Newsfeed.
I have a Post which contains all post. After sending it from the view I load the posts to the templates with {% for p in posts %}, for each post I add 2 buttons (share, like) and an input (comment).
How can I know which of them was clicked so I can send the post.id back to the view, including the name, and value and trigger a function to handle them in the database?
I want to know after I click an input what was it, what it included and for which post it belongs.

The way I would have approached this is something as follows:
I would have specific views to handle share and like functionalities, I would rather have them as two separate views. For them you then create urls, definitely you will be also providing post identifier as a parameter to this url.
For eg:
class LikePost(View):
def post(self, request, post_id, *args, **kwargs):
code
For this class, you can then have a url like ^/post/(?P<post_id>\d+)/$.
Now for more simplicity in your Post model class you can write a model method that returns you the url for LikePost view for an instance of post.
Something like this:
def like_post_url(self):
return reverse('appname:reversename-for-url', args=(self.id,))
So now when you loop posts in template you can simply assign this url as value for href in anchor tag you use for button. So every button will have its own url for share, like and also for input.
I think this should make it clear on how you can proceed with simplicity.

Related

Django - Transfer data from view to form

I am struggling with the update of database information with the forms, and simply passing information between views. I could really use some advice because I am fairly new to Django.
The flow goes like this:
1. First form; I transfer the article price and title to the view "event"
2. The view "event" handles title and price and ask for confirmation in the html form
3. Once confirmed, it directs that information to the view "transact_test", I want this view to handle the update of the database via a new form that is build with the Article model. But it provides the error message : "didn't return an HttpResponse object. It returned None instead."
To fix your error: In transact_test you are just calling render in the request.method == 'POST' block:
render(request, ...)
You need to return render:
return render(request, ...)
You should really take a look at some additional django tutorials you are making this harder than you need to. You should almost never manually render a form when using django. And as Tariq said, please don't use images.

Django run a function while exiting a view or navigating to a different URL from the current URL

In my django application, when I visit a particular URL ex:enter_database, a view function is called that adds database entries. Now when I visit a different URL, I want to clear the database entries.
My question, is it possible to call a method while leaving a view/URL.
Note: I can clear the entries by adding the logic in every other view, which is not the approach I want to do. I am looking for a way to call a method while exiting the current displayed view.
In the end of your view you have to create response object and return it.
So I don't know is a correct Django way or not, but you can create custom reponse class and insert logic inside here
class HttpResponseWithDataClearing(HttpResponse):
def __init__(self, content=b'', *args, **kwargs):
# Some custom logic here (clear the entries?)
super().__init__(content, *args, **kwargs)
After that change view's return statement
return HttpResponse(...)
↓
return HttpResponseWithDataClearing(...)
Where you want to add custom logic.
If you want to add logic when already response sent and you moving to another page, that is impossible to do at backend.
You must to set javascript action on page leaving. And do ajax request to data_clear_url
window.onunload = function() {
do_ajax_request("data_clear_url");
}
EDIT1: onunload method not working
I tried to reproduce javascript onunload method and looks like ajax request is not properly working with Chrome in this case. You can check this article

Django linking to page without putting data in URL

So I have a a HTML page with a table in it which contains details of a certain model. Each row contains details of a different object. I have a cell for a button as well.
Now, what I want is for the user to be able to click on the button and it should take them to the appropriate page for that particular user that they've clicked on. The way I can do this now is by creating a URL that takes a user_id argument along with a view to redirect it to a template. This url can then be added to the button. However, I don't want the user_id to be shown in the URL (being shown in Inspect Element is okay (as in the row ID)).
This rushed, so sorry. How can I do this?
Is there a way to do it without putting any information whatsoever in the URL?
Thank you!
One way to do this is to send user ids from a POST instead of a GET for getting the user info, when the user clicks the button, you submit a hidden form which contains user_id (which you will update accordingly) and pass it to Django. On this POST call you will wire a render of the page for the user according to the POST parameter you are expecting containing the user id.
You can read the post parameters on a request via:
request.POST.get('user_id')
The downside of this approach is that you won't be able to share the link for a certain user, because the link will only contain the get parameters.
Maybe you can refactor your application to use some kind of SPA framework on the front-end. In this way you can load any content on your current page and the URL never changes if you don't want. Take a look for example at AngularJS or Durandal. Both works well with Django.
You can also solve the problem by using POST instead of GET but in my opinion that's not a very elegant solution because POST requests should be used just when you send data to the server.
If your worried about security I don't think keeping the user_id secret will be effective but if for some other reason you have to do this put it in session and redirect to user page without any parameters.
Put your table inside a form and store the id in an attribute of the button on each row:
<button class="mybutton" data-id="{{ my_object.id }}">view</button>
Put a hidden field at the bottom of your form:
<input type="hidden" id="user_id" name="user_id" />
Javascript:
$("table .mybutton").click(function(e) {
e.preventDefault();
$("#user_id").val($(this).attr("data-id"));
$("#my_form").submit();
});
In your table view:
if request.method == "POST":
request.session["user_id"] = request.POST.get('user_id')
return redirect("user_page")
In your details view:
user_id = request.session["user_id"]
creating urls
If the url is relevant to the user; then use the user_id; e.g. http://example.com/mysite/users/<user_id>/userstuff.
obfuscation is not security.
obfuscation is not a permission scheme.
Other possibilities:
http://example.com/mysite/users/<uniqueusername>/userstuff.
http://example.com/mysite/users/<slug>/userstuff.
http://example.com/mysite/users/<encoded>/userstuff, where encoded is either 2-way encoding, or a field on the user model that is unique.
getting logged in user (request.user)
If the url has nothing to do with the user, but you need to get the authenticated user then read the docs: https://docs.djangoproject.com/en/1.7/topics/auth/default/#authentication-in-web-requests.
def my_view(request):
if request.user.is_authenticated():
# use request.user
else:
# something else

Passing template variables from URL to FormPreview in Django

I'm a Django noob and fear the answer to my question is fairly obvious, but hoping someone can help.
I'm building an app that includes the same form on every page, with the content surrounding the form and the model instance to which the form data is tied dependent on a value passed in the URL. Works fine using the standard Form class and (URL, 'template.html', myapp.view) in URLconf, like so:
url(r'^listings/(?P<listing_id>\d+)/submit$', 'myapp.views.lister'),
With FormPreview, however, instead of calling the view in the URLconf, you're calling the subclass with the view functionality baked in.
url(r'^listings/(?P<listing_id>\d+)/submit$', PickFormPreview(PickForm)),
From what I can gather from the docs, FormPreview uses parse_params to pass values captured in the URL to state.self, which I believe is a dictionary. Unfortunately given my level of experience, I can't figure out from this barebones understanding how to customize my FormPreview subclass how to pass the listing_id captured in the URL to a template variable in my form.html template called by FormPreview. Do I somehow need to override parse_params? Or somehow pass state.listing_id? Or am I missing it entirely?
Any help much appreciated!
You're on the right track. The parse_params method in FormPreview does not do anything, but it is the correct method to override in your subclass.
The following example saves listing_id to self.state.listing_id, which is available in the template context.
class PickFormPreview(FormPreview):
def parse_params(self, *args, **kwargs)
"""
Saves listing_id from the url to state
"""
self.state['listing_id'] = kwargs['listing_id']
Then in your template, you access it as:
Listing {{ state.listing_id }}

Django URL configuration

I have a purchase page, it can take an optional argument as a gift, if it is a gift, the view passes a gift form to the template and if not, a regular purchase form.
my old regular url, which redirects to two seperate views:
(r'^(?P<item>[-\w]+)/purchase/$', 'purchase_view'),
(r'^(?P<item>[-\w]+)/purchase/gift$', 'gift_view'),
and the views was like this:
def purchase_view(request,item):
....use purchase form
def gift_view(request,item):
....use giftform
It is a bad design indeed, as both views having are almost everything same but the forms used.
I have also thougt about using GET and giving gift as a GET param however it wasnt a good idea as I am using POST method for these pages, especially would cause issue after validation.
How can I make this a single url and a single view?
Thanks
urls.py
url(r'^(?P<item>[-\w]+)/purchase/$', 'purchase_view', name='purchase_view'),
url(r'^(?P<item>[-\w]+)/purchase/(?P<gift>gift)/$', 'purchase_view', name='gift_view'),
views.py
def purchase_view(request, item, gift=False):
if gift:
form = GiftForm
else:
form = PurchaseForm
...