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Django: I want to track from which url a request was made? [duplicate]

I have a Post model that requires a certain category before being added to the database, and I want the category to be generated automatically. Clicking the addPost button takes you to a different page and so the category will be determined by taking a part of the previous page URL.
Is there a way to get the previous page URL as a string?
I have added my AddPost button here.
<aside class="addPost">
<article>
<form action="/Forum/addPost">
<input type="submit" name="submit" value="Add Post"/>
</form>
</article>
</aside>

You can do that by using request.META['HTTP_REFERER'], but it will exist if only your tab previous page was from your website, else there will be no HTTP_REFERER in META dict. So be careful and make sure that you are using .get() notation instead.
# Returns None if user came from another website
request.META.get('HTTP_REFERER')
Note: I gave this answer when Django 1.10 was an actual release. I'm not working with Django anymore, so I can't tell if this applies to Django 2

You can get the referring URL by using request.META.HTTP_REFERER
More info here: https://docs.djangoproject.com/en/dev/ref/request-response/#django.http.HttpRequest.META

I can't answer #tryingtolearn comment, but for future people, you can use request.META['HTTP_REFERER']

Instead of adding it to your context, then passing it to the template, you can place it in your template directly with:
Return

A much more reliable method would be to explicitly pass the category in the URL of the Add Post button.

You can get the previous url in "views.py" as shown below:
# "views.py"
from django.shortcuts import render
def test(request):
pre_url = request.META.get('HTTP_REFERER') # Here
return render(request, 'test/index.html')
You can also get the previous url in Django Template as shown below:
# "index.html"
{{ request.META.HTTP_REFERER }}

Django-taggit tag value retrieval and formatting failing

I am trying to implement a tagging process for profiles so you can add your hobbies for example.
I have chosen django-taggit as it seemed quite simple and does what I need it to, plus don't really know how to do it myself from scratch.
I have managed to make it work to some extent but I am having issues with 3 things:
Not really sure what's the best way to control the form field for these tags as I generate the form automatically with widget adjustments in meta function of the form, but it might work fine after resolving the below two issues.
When there is no data for the field hobbies (tags) the field gets populated with a single tag of value "[]" as per below image.
When I add a tag of "music" and submit the form after I reload the page I get this "[]" as per image. I assumed this will be dealt with by the library, but I cannot see another similar scenario online.
When I try adding another tag of "games" and save and reload, the below happens. The initial value gets wrapped again.
My model is:
class UserProfile(models.Model):
user = models.OneToOneField(User, on_delete=models.CASCADE)
hobbies = TaggableManager()
My form is:
class UserProfileForm(forms.ModelForm):
class Meta:
model = UserProfile
fields = ['hobbies',]
def __init__(self, *args, **kwargs):
super(UserProfileForm, self).__init__(*args,**kwargs)
self.fields['hobbies'].widget = forms.TextInput()
self.fields['hobbies'].widget.attrs['data-role'] = "tagsinput"
self.fields['hobbies'].widget.attrs['class'] = "form-control"
self.fields['hobbies'].required = False
My view function is:
if request.method == 'POST':
user_profile = UserProfile.objects.get(user=request.user)
form = UserProfileForm(request.POST, instance=user_profile)
print(form)
if form.is_valid():
obj = form.save(commit=False)
obj.user = request.user
obj.save()
print("Form valid")
form.save_m2m()
Using:
<script src="/static/js/tagsinput.js"></script>
<link rel="stylesheet" href="{% static 'css/tagsinput.css' %}" />

I had this exact same problem.
One solution is to apply the data-role="tagsinput" AFTER you turn a list of tags into a comma-separated string for the form.
Here is that solution:
class MyModelForm(forms.ModelForm):
class Meta:
model = MyModel
def __init__(self, **kwargs):
self.fields['tags'].widget.attrs['value'] = ", ".join(list(self.instance.tags.names()))
self.fields['tags'].widget.attrs['data-role'] = "tagsinput"
Output:
As you can see, there's a problem with quotes appearing around tags that are multi-word. It also causes new tags with quotes to be saved to the database.
If double-quotes didn't appear around multi-word phrases, this would be the most elegant solution. If someone solves this in the future, drop a note!
My template is this:
<div class="m-3 p-3 border">
<form method="POST" enctype="multipart/form-data">
{% csrf_token %}
{{ form|crispy }}
<button class="btn btn-primary" type="submit">Save Form</button>
</form>
</div>
I know I can use a template tag to strip the extra quotes from the tag field itself, but then I'd have to go through and create all the form fields manually just to set the tags template tag.
For the time being, my solution is to simply use Javascript and just modify the Meta widgets section of the form.
FINAL ANSWER (for now):
forms.py
class MyModelForm(forms.ModelForm):
class Meta:
model = MyModel
widgets = {
'tags': forms.TextInput(attrs={
"data-role": "tagsinput",
})
}
custom.js - put this script on the page that loads the form.
document.addEventListener("DOMContentLoaded", function(event) {
let tags_input = document.querySelector('#id_tags');
let tags_input_value = tags_input.value;
let new_value = [...tags_input_value.matchAll(/<Tag:\s*([\w\s]+)>/g)].map(([, m]) => m).join(', ')
tags_input.setAttribute('value', new_value);
}
So all we're doing is modifying the front-end presentation, and leaving all the backend internal forms functionality untouched.

So after quite a few (hundreds) of tests, I finally narrowed down where the issue was and tried to go around it with successful result.
It seems the data got amended into tag objects through tagsinput library I was using. Only when the "data-role" was specified as "tagsinput" in the forms.py the data would already come to html side as those objects and be shown incorrectly. So instead I wanted to keep the data clean and only apply data-role='tagsinput' in the end for visual aspect, which I did using:
var hobbiesTags = document.getElementById("id_hobbies");
if(hobbiesTags){
var att = document.createAttribute("data-role");
att.value = "tagsinput";
hobbiesTags.setAttributeNode(att);
};
And that resulted in the below. Maybe there are better ways to do this, I'm not sure, but it's a pretty clean solution. Share your alternatives.

Changing Id of django form in for loop

I'm stuck in my code. Need help.
This is my front end. I am rendering forms stored in "form_list".
The problem is that the forms stored are of same type and thus produce input fields with same "id" and same "name".
This is my view:-
#login_required
def VideoLinkView(request):
"""view to save the video links """
current_form_list = []
current_form = None
if request.method == 'GET':
vl = VideoLink.objects.filter(company=CompanyModel.objects.get(owner=request.user))
for link in vl:
current_form = VideoLinkForm(link.__dict__)
current_form_list.append(current_form)
return render(request, "premium/video_link.html", context={'form_list':current_form_list})
This is my html template :-
{% for form in form_list %}
<div class="form-group">
<label for="id_video_link">Video Link:</label>
{{ form.video_link }}
</div>
{% endfor %}
How can I create different "id" and different "name" in each iteration of for loop's input tag, automatically without having knowledge of no form stored in form_list.
I tried {{ forloop.counter}} it didn't worked, perhaps I made some mistake. Also, raw python don't work in template.
Thanks in Advance.

The way you are creating your forms is wrong in two ways. Firstly, the first positional argument is for the values submitted by the user; passing this arg triggers validation, among other things. If you are passing values for display to prepopulate the form, you must use the initial kwarg:
current_form = VideoLinkForm(initial={...dict_of_values...})
However, even that is not correct for your use case here. link is a model instance; you should use the instance kwarg:
current_form = VideoLinkForm(instance=link)
Now, to solve the problem you asked, you could just pass a prefix as well as I originally recommended:
for i, link in enumerate(vl):
current_form = VideoLinkForm(instance=link, prefix="link{}".format(i))
However, now that you have shown all the details, we can see that this is not the best approach. You have a queryset; so you should simply use a model formset.
from django.forms import modelformset_factory
VideoLinkFormSet = modelformset_factory(VideoLink, form=VideoLinkForm, queryset=vl)
current_form_list = VideoLinkFormSet()

How to get the previous URL from a POST in Django

I have a Post model that requires a certain category before being added to the database, and I want the category to be generated automatically. Clicking the addPost button takes you to a different page and so the category will be determined by taking a part of the previous page URL.
Is there a way to get the previous page URL as a string?
I have added my AddPost button here.
<aside class="addPost">
<article>
<form action="/Forum/addPost">
<input type="submit" name="submit" value="Add Post"/>
</form>
</article>
</aside>

You can do that by using request.META['HTTP_REFERER'], but it will exist if only your tab previous page was from your website, else there will be no HTTP_REFERER in META dict. So be careful and make sure that you are using .get() notation instead.
# Returns None if user came from another website
request.META.get('HTTP_REFERER')
Note: I gave this answer when Django 1.10 was an actual release. I'm not working with Django anymore, so I can't tell if this applies to Django 2

You can get the referring URL by using request.META.HTTP_REFERER
More info here: https://docs.djangoproject.com/en/dev/ref/request-response/#django.http.HttpRequest.META

I can't answer #tryingtolearn comment, but for future people, you can use request.META['HTTP_REFERER']

Instead of adding it to your context, then passing it to the template, you can place it in your template directly with:
Return

A much more reliable method would be to explicitly pass the category in the URL of the Add Post button.

You can get the previous url in "views.py" as shown below:
# "views.py"
from django.shortcuts import render
def test(request):
pre_url = request.META.get('HTTP_REFERER') # Here
return render(request, 'test/index.html')
You can also get the previous url in Django Template as shown below:
# "index.html"
{{ request.META.HTTP_REFERER }}

How to implement a "back" link on Django Templates?

I'm tooling around with Django and I'm wondering if there is a simple way to create a "back" link to the previous page using the template system.
I figure that in the worst case I can get this information from the request object in the view function, and pass it along to the template rendering method, but I'm hoping I can avoid all this boilerplate code somehow.
I've checked the Django template docs and I haven't seen anything that mentions this explicitly.

Actually it's go(-1).
<input type=button value="Previous Page" onClick="javascript:history.go(-1);">

This solution worked out for me:
Go back
But that's previously adding 'django.core.context_processors.request', to TEMPLATE_CONTEXT_PROCESSORS in your project's settings.

Back
Here |escape is used to get out of the " "string.

Well you can enable:
'django.core.context_processors.request',
in your settings.TEMPLATE_CONTEXT_PROCESSORS block and hook out the referrer but that's a bit nauseating and could break all over the place.
Most places where you'd want this (eg the edit post page on SO) you have a real object to hook on to (in that example, the post) so you can easily work out what the proper previous page should be.

You can always use the client side option which is very simple:
Back

For RESTful links where "Back" usually means going one level higher:
<input type="button" value="Back" class="btn btn-primary" />

All Javascript solutions mentioned here as well as the request.META.HTTP_REFERER solution sometimes work, but both break in the same scenario (and maybe others, too).
I usually have a Cancel button under a form that creates or changes an object. If the user submits the form once and server side validation fails, the user is presented the form again, containing the wrong data. Guess what, request.META.HTTP_REFERER now points to the URL that displays the form. You can press Cancel a thousand times and will never get back to where the initial edit/create link was.
The only solid solution I can think of is a bit involved, but works for me. If someone knows of a simpler solution, I'd be happy to hear from it. :-)
The 'trick' is to pass the initial HTTP_REFERER into the form and use it from there. So when the form gets POSTed to, it passes the correct, initial referer along.
Here is how I do it:
I created a mixin class for forms that does most of the work:
from django import forms
from django.utils.http import url_has_allowed_host_and_scheme
class FormCancelLinkMixin(forms.Form):
""" Mixin class that provides a proper Cancel button link. """
cancel_link = forms.fields.CharField(widget=forms.HiddenInput())
def __init__(self, *args, **kwargs):
"""
Override to pop 'request' from kwargs.
"""
self.request = kwargs.pop("request")
initial = kwargs.pop("initial", {})
# set initial value of 'cancel_link' to the referer
initial["cancel_link"] = self.request.META.get("HTTP_REFERER", "")
kwargs["initial"] = initial
super().__init__(*args, **kwargs)
def get_cancel_link(self):
"""
Return correct URL for cancelling the form.
If the form has been submitted, the HTTP_REFERER in request.meta points to
the view that handles the form, not the view the user initially came from.
In this case, we use the value of the 'cancel_link' field.
Returns:
A safe URL to go back to, should the user cancel the form.
"""
if self.is_bound:
url = self.cleaned_data["cancel_link"]
# prevent open redirects
if url_has_allowed_host_and_scheme(url, self.request.get_host()):
return url
# fallback to current referer, then root URL
return self.request.META.get("HTTP_REFERER", "/")
The form that is used to edit/create the object (usually a ModelForm subclass) might look like this:
class SomeModelForm(FormCancelLinkMixin, forms.ModelForm):
""" Form for creating some model instance. """
class Meta:
model = ModelClass
# ...
The view must pass the current request to the form. For class based views, you can override get_form_kwargs():
class SomeModelCreateView(CreateView):
model = SomeModelClass
form_class = SomeModelForm
def get_form_kwargs(self):
kwargs = super().get_form_kwargs()
kwargs["request"] = self.request
return kwargs
In the template that displays the form:
<form method="post">
{% csrf token %}
{{ form }}
<input type="submit" value="Save">
Cancel
</form>

For a 'back' button in change forms for Django admin what I end up doing is a custom template filter to parse and decode the 'preserved_filters' variable in the template. I placed the following on a customized templates/admin/submit_line.html file:
<a href="../{% if original}../{% endif %}?{{ preserved_filters | decode_filter }}">
{% trans "Back" %}
</a>
And then created a custom template filter:
from urllib.parse import unquote
from django import template
def decode_filter(variable):
if variable.startswith('_changelist_filters='):
return unquote(variable[20:])
return variable
register = template.Library()
register.filter('decode_filter', decode_filter)

Using client side solution would be the proper solution.
Cancel