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.
Be forewarned of a triple-newbie threat - new to python, new to python anywhere, new to flask.
[pythonanywhere-root]/mysite/test01.py
# A very simple Flask Hello World app for you to get started with...
from flask import Flask
from flask import render_template # for templating
#from flask import request # for handling requests eg form post, etc
app = Flask(__name__)
app.debug = True #bshark: turn on debugging, hopefully?
#app.route('/')
#def hello_world():
# return 'Hello from Flask! wheee!!'
def buildOrg():
orgname = 'ACME Inc'
return render_template('index.html', orgname)
And then in [pythonanywhere-root]/templates/index.html
<!doctype html>
<head><title>Test01 App</title></head>
<body>
{% if orgname %}
<h1>Welcome to {{ orgname }} Projects!</h1>
{% else %}
<p>Aw, the orgname wasn't passed in successfully :-(</p>
{% endif %}
</body>
</html>
When I hit up the site, I get 'Unhandled Exception' :-(
How do I get the debugger to at least spit out where I should start looking for the problem?
The problem is render_template only expects one positional argument, and rest of the arguments are passed as keyword only arguments.So, you need to change your code to:
def buildOrg():
orgname = 'ACME Inc'
return render_template('index.html', name=orgname)
For the first part, you can find the error logs under the Web tab on pythonanywhere.com.
You need to also pass your name of orgname variable that is used in your template to render_template.
flask.render_template:
flask.render_template(template_name_or_list, **context)
Renders a template from the template folder with the given context.
Parameters:
template_name_or_list – the name of the template to be rendered,
or an iterable with template names the first one existing will be rendered
context – the variables that should be available in the context of the template.
So, change this line:
return render_template('index.html', orgname)
To:
return render_template('index.html', orgname=orgname)
I have a problem while using jinja2 url_for() function.
I have a route like this:
#app.route('/article/<int:article_id>/<url_title>/', methods=['GET'])
def article_page(article_id, url_title):
article = Article.query.get(article_id)
if article == None:
abort(404)
return render_template('article.html', article=article)
in jinja template file,i want to create a url which links to article_page,so i write like this:
<h5>
{{ article.title }}
</h5>
but when I run this page, I get a error:
raise BuildError(endpoint, values, method)
BuildError: ('article_page', {'article_id': 1}, None)
It seems like that the second parameter url_title missing.
How can I use url_for() with multiple parameters correctly?
According to url_for documentation:
If the value of a query argument is None, the whole pair is skipped.
Make sure that url_title is not None.
Or specify default value for url_title in the article_page function.
You missed a dot, try that:
url_for('.article_page', article_id=article.id, url_title=article.url_title)
Hello? I am trying to redirect a particular user to a custom page once they log in in django. The Admin will be directed to their usual admin interface while this particular user will go to their own custom page.
I have written this code and placed it in my views
def custLogin(request):
if request.user.username == '***':
return HttpResponseRedirect('http://********************.html')
else:
return login(request,template_name='login.html')
I have pointed the accounts/login url in urls.py to custLogin as below
(r'^accounts/login/', custLogin),
I however keep getting the error
Caught NoReverseMatch while rendering: Reverse for 'django.contrib.auth.views.login' with arguments '()' and keyword arguments '{}' not found.
Any pointers please?
Maybe, you should use redirect shortcut, which returns an HttpResponseRedirect, to point your users to different places after custLogin view processed.
from django.shortcuts import redirect
def custLogin(request):
if request.user.is_staff:
return redirect('/some/admin/url')
else:
return redirect('/some/regular/user/url')
As documentation says, you can use the redirect() function in a number of ways: by passing a view name, an object or url. See documentation for more info
Note, that I use is_staff user field to determine: is user an admin or not.
For redirecting each user to his own page use reverse
First create url route which accepts parameter. It's described here
And redirect like
return HttpResponseRedirect(reverse('arch-summary', args=[1945]))
where arch-summary - named url, and in list args - parameters for it. Example here
To distinguish admin user from usual user, check it in view and redirect to different url.
In your template, you may have forgotten to do this at the beginning of your template:
{% load from future url %}
This is useful when using:
{% url 'name_of_my_url_inurl.py' %}
How can I get the full/absolute URL (e.g. https://example.com/some/path) in Django without the Sites module? That's just silly... I shouldn't need to query my DB to snag the URL!
I want to use it with reverse().
Use handy request.build_absolute_uri() method on request, pass it the relative url and it'll give you full one.
By default, the absolute URL for request.get_full_path() is returned, but you can pass it a relative URL as the first argument to convert it to an absolute URL.
>>> request.build_absolute_uri()
'https://example.com/music/bands/the_beatles/?print=true'
>>> request.build_absolute_uri('/bands/?print=true')
'https://example.com/bands/?print=true'
If you want to use it with reverse() you can do this : request.build_absolute_uri(reverse('view_name', args=(obj.pk, )))
If you can't get access to request then you can't use get_current_site(request) as recommended in some solutions here. You can use a combination of the native Sites framework and get_absolute_url instead. Set up at least one Site in the admin, make sure your model has a get_absolute_url() method, then:
>>> from django.contrib.sites.models import Site
>>> domain = Site.objects.get_current().domain
>>> obj = MyModel.objects.get(id=3)
>>> path = obj.get_absolute_url()
>>> url = 'http://{domain}{path}'.format(domain=domain, path=path)
>>> print(url)
'http://example.com/mymodel/objects/3/'
https://docs.djangoproject.com/en/dev/ref/contrib/sites/#getting-the-current-domain-for-full-urls
You can also use get_current_site as part of the sites app (from django.contrib.sites.models import get_current_site). It takes a request object, and defaults to the site object you have configured with SITE_ID in settings.py if request is None. Read more in documentation for using the sites framework
e.g.
from django.contrib.sites.shortcuts import get_current_site
request = None
full_url = ''.join(['http://', get_current_site(request).domain, obj.get_absolute_url()])
It isn't as compact/neat as request.build_absolute_url(), but it is usable when request objects are unavailable, and you have a default site url.
If you don't want to hit the database, you could do it with a setting. Then, use a context processor to add it to every template:
# settings.py (Django < 1.9)
...
BASE_URL = 'http://example.com'
TEMPLATE_CONTEXT_PROCESSORS = (
...
'myapp.context_processors.extra_context',
)
# settings.py (Django >= 1.9)
TEMPLATES = [
{
'BACKEND': 'django.template.backends.django.DjangoTemplates',
'DIRS': [],
'APP_DIRS': True,
'OPTIONS': {
'context_processors': [
'django.template.context_processors.debug',
'django.template.context_processors.request',
'django.contrib.auth.context_processors.auth',
'django.contrib.messages.context_processors.messages',
# Additional
'myapp.context_processors.extra_context',
],
},
},
]
# myapp/context_processors.py
from django.conf import settings
def extra_context(request):
return {'base_url': settings.BASE_URL}
# my_template.html
<p>Base url is {{ base_url }}.</p>
In your view, just do this:
base_url = "{0}://{1}{2}".format(request.scheme, request.get_host(), request.path)
This worked for me in my template:
{{ request.scheme }}://{{ request.META.HTTP_HOST }}{% url 'equipos:marca_filter' %}
I needed the full url to pass it to a js fetch function.
I hope this help you.
django-fullurl
If you're trying to do this in a Django template, I've released a tiny PyPI package django-fullurl to let you replace url and static template tags with fullurl and fullstatic, like this:
{% load fullurl %}
Absolute URL is: {% fullurl "foo:bar" %}
Another absolute URL is: {% fullstatic "kitten.jpg" %}
These badges should hopefully stay up-to-date automatically:
In a view, you can of course use request.build_absolute_uri instead.
Yet another way. You could use build_absolute_uri() in your view.py and pass it to the template.
view.py
def index(request):
baseurl = request.build_absolute_uri()
return render_to_response('your-template.html', { 'baseurl': baseurl })
your-template.html
{{ baseurl }}
Examine Request.META dictionary that comes in. I think it has server name and server port.
To create a complete link to another page from a template, you can use this:
{{ request.META.HTTP_HOST }}{% url 'views.my_view' my_arg %}
request.META.HTTP_HOST gives the host name, and url gives the relative name. The template engine then concatenates them into a complete url.
Try the following code:
{{ request.scheme }}://{{ request.META.HTTP_HOST }}
If you're using django REST framework, you can use the reverse function from rest_framework.reverse. This has the same behavior as django.core.urlresolvers.reverse, except that it uses a request parameter to build a full URL.
from rest_framework.reverse import reverse
# returns the full url
url = reverse('view_name', args=(obj.pk,), request=request)
# returns only the relative url
url = reverse('view_name', args=(obj.pk,))
Edited to mention availability only in REST framework
I know this is an old question. But I think people still run into this a lot.
There are a couple of libraries out there that supplement the default Django functionality. I have tried a few. I like the following library when reverse referencing absolute urls:
https://github.com/fusionbox/django-absoluteuri
Another one I like because you can easily put together a domain, protocol and path is:
https://github.com/RRMoelker/django-full-url
This library allows you to simply write what you want in your template, e.g.:
{{url_parts.domain}}
If anyone is interested in fetching the absolute reverse url with parameters in a template , the cleanest way is to create your own absolute version of the {% url %} template tag by extending and using existing default code.
Here is my code:
from django import template
from django.template.defaulttags import URLNode, url
register = template.Library()
class AbsURLNode(URLNode):
def __init__(self, view_name, args, kwargs, asvar):
super().__init__(view_name, args, kwargs, asvar)
def render(self, context):
url = super().render(context)
request = context['request']
return request.build_absolute_uri(url)
#register.tag
def abs_url(parser, token):
urlNode = url(parser, token)
return AbsURLNode( urlNode.view_name, urlNode.args, urlNode.kwargs, urlNode.asvar )
Usage in templates:
{% load wherever_your_stored_this_tag_file %}
{% abs_url 'view_name' parameter %}
will render(example):
http://example.com/view_name/parameter/
instead of
/view_name/parameter/
I got it:
wsgiref.util.request_uri(request.META)
Get the full uri with schema, host, port path and query.
You can either pass request reverse('view-name', request=request) or enclose reverse() with build_absolute_uri request.build_absolute_uri(reverse('view-name'))
Not for absolute url but I was looking just to get host. If you want to get host in your view.py you can do
def my_view(request):
host = f"{ request.scheme }://{ request.META.get('HTTP_HOST') }"
As mentioned in other answers, request.build_absolute_uri() is perfect if you have access to request, and sites framework is great as long as different URLs point to different databases.
However, my use case was slightly different. My staging server and the production server access the same database, but get_current_site both returned the first site in the database. To resolve this, you have to use some kind of environment variable. You can either use 1) an environment variable (something like os.environ.get('SITE_URL', 'localhost:8000')) or 2) different SITE_IDs for different servers AND different settings.py.
Hopefully someone will find this useful!
While working on a project I came to know to get the full/absolute URL in Django.
If your URL looks like this in the address bar:
https://stackoverflow.com/questions/2345708
And if you want to show the above URL to your template.
{{ request.path }} #Without GET parameters.
{{ request.get_full_path }} #with GET parameters
For the above two codes, this will print in your template will be
questions/2345708
and another way to get a full URL is:
{{request.build_absolute_uri}}
this will print in your template will be:
https://stackoverflow.com/questions/2345708
There is also ABSOLUTE_URL_OVERRIDES available as a setting
https://docs.djangoproject.com/en/2.1/ref/settings/#absolute-url-overrides
But that overrides get_absolute_url(), which may not be desirable.
Instead of installing sites framework just for this or doing some of the other stuff mentioned here that relies on request object, I think the better solution is to place this in models.py
Define BASE_URL in settings.py, then import it into models.py and make an abstract class (or add it to one you're already using) that defines get_truly_absolute_url(). It could be as simple as:
def get_truly_absolute_url(self):
return BASE_URL + self.get_absolute_url()
Subclass it and now you can use it everywhere.
I came across this thread because I was looking to build an absolute URI for a success page. request.build_absolute_uri() gave me a URI for my current view but to get the URI for my success view I used the following....
request.build_absolute_uri(reverse('success_view_name'))
<div class='col-12 col-md-6'>
<p class='lead'>Login</p>
{% include 'accounts/snippets/form.html' with form=login_form next_url=request.build_absolute_uri %}
</div>
Here for example am saying load the form and tell the form that the next URL is the current URL which this code rendred from
I use this code :
request.build_absolute_uri('/')[:-1]
response :
https://yourdomain.com
request.get_host() will give you the domain.
request.get_host()
Use this for request object for APIView in django
class WalletViewSet(mixins.ListModelMixin, GenericViewSet):
serializer_class = WalletSerializers
pagination_class = CustomPaginationInvestment
def get_queryset(self):
######################################################
print(self.request.build_absolute_uri())
#####################################################
wallet, created = Wallet.objects.get_or_create(owner=self.request.user)
return Wallet.objects.filter(id=wallet.id)
You get output like this
http://localhost:8000/v1/wallet
HTTP GET /v1/wallet 200 [0.03, 127.0.0.1:41608]
You can also use:
import socket
socket.gethostname()
This is working fine for me,
I'm not entirely sure how it works. I believe this is a bit more low level and will return your server hostname, which might be different than the hostname used by your user to get to your page.
You can try "request.get_full_path()"