Customizing django-allauth password_reset_from_key template - python

I'm using django 2.2.3, I'm overriding django-allauth templates and on password_reset_from_key template I came across a NoReverseMatch Exception,
Reverse for 'account_reset_password_from_key' with no arguments not found. 1 pattern(s) tried: ['accounts/password/reset/key/(?P<uidb36>[0-9A-Za-z]+)-(?P<key>.+)/$']
So it expects argument and probably that argument is the key that is generated during resetting password. From the URL pattern, 'key' is the variable's name
I tried to pass it with 'key'.
<form method="POST" action="{% url 'account_reset_password_from_key' key %}">
...
</form>
but it did not work. What is wrong?

If you're not overriding the allauth view, then use action_url:
<form method="POST" action="{{ action_url }}">
that's what the view passes to the template.

You didn't supply where that error is coming from, but if I look for where that url name is used in the code, it's not being used in a template.
See:
https://github.com/pennersr/django-allauth/search?q=account_reset_password_from_key&unscoped_q=account_reset_password_from_key
My guess is you're calling this url name and not passing the parameters to it?
url(r"^password/reset/key/(?P<uidb36>[0-9A-Za-z]+)-(?P<key>.+)/$",
views.password_reset_from_key,
name="account_reset_password_from_key"),
PS:
As a general rule when overriding django app templates I'd suggest starting with the original template (eg copy the file allauth/templates/account/password_reset.html or whatever to your templates dir (creating all the subdirectories)), and then change the pieces you need - that way if something goes wrong it's easy to roll back and figure out why.
I know from experience that allauth templates are fairly complex and it's easy to break functionality in them if you don't fully understand how they work.

Related

Cannot have any URLs with slugs. NoReverseMatch

I'm a begginer grasping at straws with difficulty dealing with the django slug url system and these NoReverseMatch errors that make no sense to me even after reading the docs.
I have a django project. In one of the views, I pass a list of geoJSON features into a template, and show them on a map. I want to have each feature act as a clickable 'link' to a view that will show stuff about it. The following is part of the template that has those features that I want to click on:
//part of the template:
<script type="text/javascript">
...
function onEachFeature(feature, layer) {
layer.on('click', function (e) {
window.location.href = "{% url 'polls:areadetail' feature.properties.myslug%}";
});
}
(I have confirmed that feature.properties.myslug does in fact contain the slug I want).
The url pattern I want to go to:
urlpatterns = [...
url(r'^areadetail/(?P<areaslug>[-\w]+)/$', views.AreaDetail, name='areadetail'),]
And the view it relates to:
def AreaDetail(request, areaslug):
area = get_object_or_404(Area, nameslug=areaslug)
return render(request, 'polls/areadetail.html', {'area':area})
The issue I get is, by doing what I show and placing that url reference inside that template I show above, that I want to be able click on, that template won't even work at all, giving me a 'Error during template rendering' full page error info that starts with:
NoReverseMatch at /polls/areas/
Reverse for 'areadetail' with arguments '('',)' and keyword arguments '{}' not found. 1 pattern(s) tried: [u'polls/areadetail/(?P[-\w]+)/$']
Any help would be immensely appreciated
EDIT part1: As I've said in response to falsetru, I'm sure feature.properties.myslug has in fact got a slug expression in it.
EDIT2: Based on something I found in a django ticket, I've made a slight change in the url regex at urls.py, from (?P<areaslug>[-\w]+)/$ to (?P<areaslug>[-\w]+)?/$ and now the error is:
Page not found (404)
Request Method: GET Request URL: http://127.0.0.1:8000/polls/areadetail// Raised by: polls.views.AreaDetail
Is it possible that because the "{% url 'polls:areadetail' feature.properties.myslug%}" bit is inside javascript, that feature.properties.myslug is not being inserted there correctly? Like some sort of brackets are needed here?
According to the error message, feature.properties.myslug is empty or has no value.
Make sure the feature.properties.myslug is passed correctly from view.
Comment out {% url .. %} temporarily.
Print {{ feature }}, {{ feature.properties }}, {{ feature.properties.myslug }} to see if which part is missing.
Fix view accordingly.
Uncomment {% url .. %}.
After some more digging around I've found the answer to why doesn't this work in another question at:
How to pass javascript variable to django custom filter
The answer to it by Ludwik Trammer says:
Django templates are build on the server side, while JavaScript is executed on the client side.
That means that template code is always executed before JavaScript (as
it is executed by the server, before the page is sent to the client).
As a consequence it is absolutely impossible to mix JavaScript and
Django code the way you want to.
Which clearly applies here. I was focused on problems with the URL template, regex on the urls.py file etc. when the problem was that no matter what I did, because it's in a javascript section, run client-side, that URL template will always be incomplete no matter what I do, therefore being an impossible solution to what I want.

Pass <input> value to Django view

Let's say I have the following pointless example view:
def foo(request, input):
return HttpResponse()
and in a template I have a form:
<form method="get" action="{% url 'foo' ??? %}">
<input id="myinput" type="text" name="myinput">
...
</form>
Finally, I have the following url in my URLconf:
urlpatterns = [
url(r'^foo/(.+)/', views.foo, name='foo'),
]
What I would like to do, is pass the value entered by the user into the input with the id of #myinput to the foo() view function. To put it another way, you should be able to enter bar in the html input, and when you submit the form it will take you to foo/bar/.
I know that within the foo view I could access the value of the input easily with request.GET['myinput'], but I want it to show up in the url as well.
This seems like it should be a fairly common task, but I have not been able to come up with a solution yet. Any suggestions would be appreciated. My Frankenstein's Monster of a first Django site is almost complete, and this is one of last pieces I am missing.
The source of my misunderstanding
Although I did not make this clear in an attempt to simplify my example and avoid using app-specific code, my use case is a simple search view. The view was actually one of the first views I wrote in the start of my Django journey, and I mistakenly was POSTing my data instead of GETing it. This was making it so that if I was searching for the item foo, it would take me to the detail page for foo, but the url would be mysite/search/ (i.e., the search query is not included in the url though it is included in the request), and I can't return to those search results by visiting the url mysite/search/.
While I was using a GET request in my toy example in this question, I didn't realize that I had been using a POST in my app, and that with some minor tweaking I can get the functionality I want for free very easily. I know that all of this is extremely obvious to veteran and even intermediate web developers, but for someone starting from scratch without web or cs experience, things like HTTP can be a little confusing. At least for me it is. Thanks so much to #Two-Bit Alchemist for explaining this in a way that I can understand.
Applying all this to my toy example
I would get rid of the passed parameter in my view:
def foo(request):
# If I want to do something with the search query, I can access it with
# request.GET['search_query']
return HttpResponse()
change my form in my template to:
<form method="get" action="{% url 'foo' %}">
<input id="myinput" type="text" name="search_query">
...
</form>
and change my url to:
urlpatterns = [
url(r'^foo/search/', views.foo, name='foo'),
]
As #Two-Bit Alchemist said: "The rest will happen like magic". If a user enters bar in the input and submits the form, they will be taken to foo/search/?search_query=bar. This is what I was looking for.

Unable to access request.GET['next'] in get_login_redirect_url in Django All Auth

My form submits as follows
<form class="form-signin" role="form" action="{% provider_login_url "facebook" method="js_sdk" next="/next"%}">
I overrode the DefaultAccountAdapter with my own AccountAdapter with method
def get_login_redirect_url(self, request):
print request.GET['next']
...
But request loses the next parameter and the print returns an error because there is no "next" in request.GET.
Why can't I access the next parameter?
I was originally using get_login_redirect_url to handle different url redirects after creation of social versus username/password users. Now, I need to be able to specific the next parameter in the URL for another variant of behavior for social user login but am unable to access the next parameter because it does not seem to be passed.
I am not sure whether I could give the precise solution for your issue. But I think got the point.
To access the next parameter from url,
The url should be,
http://127.0.0.1:8000/index?next=2
If you have to form the url in this above manner,you can get access to the next argument from request object in your corresponding view method
print request.GET.get('next')
So, please make sure to format request url with proper querystring refer
To your case,
I have no idea about {% provider_login_url %} template tag
I am assuming after your tag rendered it yields the url index, then i am appending my querystring next
<form class="form-signin" role="form" action="/index?next=someValue">
you may try additionally,
{% provider_login_url "facebook" next=next %}
source
I came across with a similar problem. However, I wasn't signing in with facebook. request.GET was always empty.
I think you could try using jQuery to manually append next parameter to the action attribute of <form>. See this question. It solves my problem.

Django url template confusion

Okay I am having a bit of an issue.
I want to create a button with a link, and right now I am using action={% url views.contest_overview %} in hopes that the reverse lookup by Django will match (r'^category/$', views.contest_overview), in my urls.py. However, this is not working and I can't figure out the proper nomenclature, despite numerous guesses.
The error I get (with my best guess above) is:
Caught NoReverseMatch while rendering: Reverse for
'views.contest_overview' with arguments '()' and keyword arguments
'{}' not found.
Thank you very much for your time!
Use the application name in the url tag, e.g. {% url myapp.views.contest_overview %}
This is what I usually do; I give names to my url. For example:
url(r'^account/register/$', 'someapp.views.register_view', name='account_register'),
Therefore in template, I can do this:
{% url account_register as url_acc_register %}
<html>
..
..
Some link

How do I add data to an existing model in Django?

Currently, I am writing up a bit of a product-based CMS as my first project.
Here is my question. How can I add additional data (products) to my Product model?
I have added '/admin/products/add' to my urls.py, but I don't really know where to go from there. How would i build both my view and my template? Please keep in mind that I don't really know all that much Python, and i am very new to Django
How can I do this all without using this existing django admin interface.
You will want to wire your URL to the Django create_object generic view, and pass it either "model" (the model you want to create) or "form_class" (a customized ModelForm class). There are a number of other arguments you can also pass to override default behaviors.
Sample URLconf for the simplest case:
from django.conf.urls.defaults import *
from django.views.generic.create_update import create_object
from my_products_app.models import Product
urlpatterns = patterns('',
url(r'^admin/products/add/$', create_object, {'model': Product}))
Your template will get the context variable "form", which you just need to wrap in a <form> tag and add a submit button. The simplest working template (by default should go in "my_products_app/product_form.html"):
<form action="." method="POST">
{{ form }}
<input type="submit" name="submit" value="add">
</form>
Note that your Product model must have a get_absolute_url method, or else you must pass in the post_save_redirect parameter to the view. Otherwise it won't know where to redirect to after save.
This topic is covered in Django tutorials.
Follow the Django tutorial for setting up the "admin" part of an application. This will allow you to modify your database.
Django Admin Setup
Alternatively, you can just connect directly to the database using the standard tools for whatever database type you are using.

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