I'm working with web.py on a project with a user authentication component. This project has a /demo directory and a /register directory.
The problem is that for some reason, clicking the button meant to travel to the /register directory goes to /demo instead. Interestingly, if I type /register at the end of the URL, /register appears properly. The URL is different in this case, lacking the ?register=Register part that comes from clicking the button.
The specific error that appears when I click on the register button is this:
type 'exceptions.TypeError'> at /demo
__template__() takes exactly 1 argument (0 given)
The code that it points to as a problem area is this:
class demo:
def GET(self):
render = create_render(session.get('privilege', 0))
return '%s' % render.demo() # Specifically this line generates the error.
I don't think there is any weird fall through, as in the code there are two classes implemented between the register class and the demo class. They're actually implemented in the order they're displayed in the urls list, with reset and profile in between the two.
I don't think there is anything wrong with my urls list:
urls = (
'/', 'index',
'/register', 'register',
'/reset', 'reset',
'/profile', 'profile',
'/demo', 'demo',
'/entropy', 'entropy',
)
Here also is the HTML code for the buttons:
<form action="demo" method="GET">
<input type="submit" name="demo" value="Demo"/>
<form/>
<form action="register" method="GET">
<input type="submit" name="register" value="Register"/>
</form>
Other potentially relevant bits of code:
register.GET():
class register:
def GET(self):
reg_form = forms.registration_form()
render = create_render(session.get('privilege'))
return render.register()
create_privilege(privilege): (Explained here, at #4)
def create_render(privilege):
if logged():
render = web.template.render('templates/logged', base='base')
else:
render = web.template.render('templates/', base="base")
return render
So, I suppose, the final question is this: Why is the button rendering the wrong page? I can post more code if need be!
Thank you!
I fixed it! There was an issue in one of the HTML buttons, a form tag was not closed properly. What a silly error.
Related
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 }}
I have been using django-filters in many of my projects where the search form is on the same page as the list of results and that works just fine, but now im faced by a scenario where i have a search box in my home page and a results page somewhere else, how can i pass the filtered results to the view of my results page?
You could go a couple of different ways.
One idea is to just render a different template if there is search data present in the request.
Seeing as you didn't post any of your own code, we don't know how you have things named or set-up so here is a quick sketch of the kind of thing I mean.
class SomeView(View):
template_name = 'some_view.html'
results_template_name = 'some_view_results.html'
def get(self, request, *args, **kwargs):
if not request.GET.get('q'):
return render(request, self.template_name)
results = SearchFilter(request.GET)
context = {'results': results}
return render(request, self.results_template_name, context)
This would be one of the simplest solutions. You could even get away with using one template and just conditionally dispatch some rendering over whether or not results is in the context.
Another would be to do a full redirect and pass along whatever data you need as kwargs to whatever view you are calling. But that is kind of messy and unnecessary.
The best solution is to wire up a new ResultsView endpoint, move the filtering to that view, and put a reference that URL into the form node on the template. Which would require something like this.
<form method="GET" action="{% url search %}">
<input type="text" id="search" name="q" placeholder="Search">
</form>
Assuming that you have named ResultsView as search in your URL config.
For example, now if I have two buttons in a form element, when you click on either one of them, you'll be directed to the corresponding profile.
<form action="{{ url_for('getProfile') }}" method="post">
<button type="submit" name="submit" value="profile1"> View Profile</button>
<button type="submit" name="submit" value="profile2"> View Profile</button>
</form>
In my apprunner.py, I have
#app.route('/profile', methods=['POST'])
def getProfile():
if request.form['submit'] = 'profile1':
return render_template("profile1.html")
else if request.form['submit'] = 'profile2':
return render_template("profile2.html")
However, my problem is when I click on either button, the url will always be something like "127.0.0.1:5000/profile". But, I want it to look like "http://127.0.0.1:5000/profile1" or "http://127.0.0.1:5000/profile2".
I have looked for solution on how to generate dynamic URL online, but none of them works for button click.
Thanks in advance!
#app.route('/profile<int:user>')
def profile(user):
print(user)
You can test it on a REPL:
import flask
app = flask.Flask(__name__)
#app.route('/profile<int:user>')
def profile(user):
print(user)
ctx = app.test_request_context()
ctx.push()
flask.url_for('.profile', user=1)
'/profile1'
EDIT:
how you pass the user parameter to your new route depends on what you need. If you need hardcoded routes for profile1 and profile2 you can pass user=1 and user=2 respectively. If you want to generate those links programatically, depends on how these profiles are stored.
Otherwise you could redirect instead of render_template, to the url_for with the parsed element in the request object. This means having two routes
#app.route('/profile<int:user>')
def profile_pretty(user):
print(user)
#app.route('/profile', methods=['POST'])
def getProfile():
if request.form['submit'] = 'profile1':
return redirect(url_for('.profile_pretty', user=1))
else if request.form['submit'] = 'profile2':
return redirect(url_for('.profile_pretty', user=2))
caveat: This would make your routes look like you want, but this is inefficient as it generates a new request each time, just to make your urls the way you want. At this point it's safe to ask why do you want to have dynamically generated routes for static content.
As explained in http://exploreflask.com/en/latest/views.html#url-converters
When you define a route in Flask, you can specify parts of it that will be converted into Python variables and passed to the view function.
#app.route('/user/<username>')
def profile(username):
pass
Whatever is in the part of the URL labeled will get passed to the view as the username argument. You can also specify a converter to filter the variable before it’s passed to the view.
#app.route('/user/id/<int:user_id>')
def profile(user_id):
pass
In this code block, the URL http://myapp.com/user/id/Q29kZUxlc3NvbiEh will return a 404 status code – not found. This is because the part of the URL that is supposed to be an integer is actually a string.
We could have a second view that looks for a string as well. That would be called for /user/id/Q29kZUxlc3NvbiEh/ while the first would be called for /user/id/124.
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 }}
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