Before Django 1.0 there was an easy way to get the admin url of an object, and I had written a small filter that I&#39;d use like this: `&lt;a href=&quot;{{ object|admin_url }}&quot; .... &gt; ... &lt;/a&gt;`

Basically I was using the url reverse function with the view name being `&#39;django.contrib.admin.views.main.change_stage&#39;`

    reverse( &#39;django.contrib.admin.views.main.change_stage&#39;, args=[app_label, model_name, object_id] )

to get the url.

As you might have guessed, I&#39;m trying to update to the latest version of Django, and this is one of the obstacles I came across, that method for getting the admin url doesn&#39;t work anymore.

How can I do this in django 1.0? (or 1.1 for that matter, as I&#39;m trying to update to the latest version in the svn).



Getting Django admin url for an object

I have a Django URL like this:


    url(
        r&#39;^project_config/(?P&lt;product&gt;\w+)/(?P&lt;project_id&gt;\w+)/$&#39;,
        &#39;tool.views.ProjectConfig&#39;,
        name=&#39;project_config&#39;
    ),


**views.py:**

    def ProjectConfig(request, product, project_id=None, template_name=&#39;project.html&#39;):
        ...
        # do stuff

The problem is that I want the `project_id` parameter to be optional.

I want `/project_config/` and `/project_config/12345abdce/` to be equally valid URL patterns, so that **if** `project_id` is passed, **then** I can use it.

As it stands at the moment, I get a 404 when I access the URL without the `project_id` parameter.

Django optional url parameters

How can I see the current urlpatterns that &quot;reverse&quot; is looking in? 

I&#39;m calling reverse in a view with an argument that I think should work, but doesn&#39;t. Any way I can check what&#39;s there and why my pattern isn&#39;t?

Django : How can I see a list of urlpatterns?

In a django online course, the instructor has us use the `url()` function to call views and utilize regular expressions in the urlpatterns list. I&#39;ve seen other examples on youtube of this.
e.g.

    from django.contrib import admin
    from django.urls import include
    from django.conf.urls import url
    
    urlpatterns = [
        path(&#39;admin/&#39;, admin.site.urls),
        url(r&#39;^polls/&#39;, include(&#39;polls.urls&#39;)),
    ]


    #and in polls/urls.py

    urlpatterns = [        
        url(r&#39;^$&#39;, views.index, name=&quot;index&quot;),
    ]

However, in going through the Django tutorial, they use `path()` instead e.g.:

    from django.urls import path
    from . import views
    
    urlpatterns = [
        path(&#39;&#39;, views.index, name=&quot;index&quot;),        
    ]

Furthermore regular expressions don&#39;t seem to work with the `path()` function as using a `path(r&#39;^$&#39;, views.index, name=&quot;index&quot;)` won&#39;t find the `mysite.com/polls/` view. 

Is using `path()` without regex matching the proper way going forward? Is `url()` more powerful but more complicated so they&#39;re using `path()` to start us out with? Or is it a case of different tools for different jobs?

Is it better to use path() or url() in urls.py for django 2.0?

I am confused by `static root` and want to clarify things.

To serve static files in Django, the following should be in `settings.py` and `urls.py`:

    import os
    PROJECT_DIR=os.path.dirname(__file__)

# 1. Absolute path to the directory in which static files should be collected

    STATIC_ROOT= os.path.join(PROJECT_DIR,&#39;static_media/&#39;)

# 2. URL prefix for static files

    STATIC_URL = &#39;/static/&#39;

# 3. Additional locations for static files

    STATICFILES_DIRS = ( os.path.join(PROJECT_DIR,&#39;static/&#39;),)

...and in `urls.py` the following lines:

    from django.contrib.staticfiles.urls import staticfiles_urlpatterns
    urlpatterns += patterns(&#39;&#39;, (
        r&#39;^static/(?P&lt;path&gt;.*)$&#39;,
        &#39;django.views.static.serve&#39;,
        {&#39;document_root&#39;: settings.STATIC_ROOT}
    ))

# 4. We also use `python manage.py collectstatic`

Questions:

1. Could anyone please explain the workflow to me: how should things ideally be done. As of now, I copy/paste the above code snippets into their designated locations and continue making new files in the static directory and it works. In my `settings.STATIC_ROOT`, however, I have pointed to a different directory. 

2. It would be great if someone could explain the workflow of each setting: how files are collected and managed, and what would be a good practice to follow. 
 
Thanks. 

Difference between static STATIC_URL and STATIC_ROOT on Django

I have some code and when it executes, it throws a NoReverseMatch, saying:

&gt;NoReverseMatch at /my_url/ Reverse for &#39;my_url_name&#39; with arguments &#39;()&#39; and keyword arguments &#39;{}&#39; not found. n pattern(s) tried: []

What does this mean, and what can I do about it?

What is a NoReverseMatch error, and how do I fix it?

I am currently trying out Django. I use the ``namespace`` argument in one of my ``include()``s in urls.py. When I run the server and try to browse,
I get this error.

    File &quot;C:\Users\User\AppData\Local\Programs\Python\Python36-32\lib\site-packages\django\urls\conf.py&quot;, line 39, in include
        &#39;Specifying a namespace in include() without providing an app_name &#39;
    django.core.exceptions.ImproperlyConfigured: Specifying a namespace in include() without providing an app_name is not supported. Set the app_name attribute in the included module, or pass a 2-tuple containing the list of patterns and app_name instead.

These are my urls.py files:

    #project/urls.py

    from django.conf.urls import include, url
    from django.contrib import admin
    
    urlpatterns = [
        url(r&#39;^reviews/&#39;, include(&#39;reviews.urls&#39;, namespace=&#39;reviews&#39;)),
        url(r&#39;^admin/&#39;, include(admin.site.urls)),
    ]

and

    #app/urls.py

    from django.conf.urls import url
    
    from . import views
    
    urlpatterns = [
        # ex: /
        url(r&#39;^$&#39;, views.review_list, name=&#39;review_list&#39;),
        # ex: /review/5/
        url(r&#39;^review/(?P&lt;review_id&gt;[0-9]+)/$&#39;, views.review_detail, name=&#39;review_detail&#39;),
        # ex: /wine/
        url(r&#39;^wine$&#39;, views.wine_list, name=&#39;wine_list&#39;),
        # ex: /wine/5/
        url(r&#39;^wine/(?P&lt;wine_id&gt;[0-9]+)/$&#39;, views.wine_detail, name=&#39;wine_detail&#39;),
    ]

What do I pass the ``app_name`` as stated in the error message?


ImproperlyConfiguredError about app_name when using namespace in include()

How can I redirect traffic that doesn&#39;t match any of my other URLs back to the home page?

**urls.py:**

    urlpatterns = patterns(&#39;&#39;,
        url(r&#39;^$&#39;, &#39;macmonster.views.home&#39;),
        #url(r&#39;^macmon_home$&#39;, &#39;macmonster.views.home&#39;),
        url(r&#39;^macmon_output/$&#39;, &#39;macmonster.views.output&#39;),
        url(r&#39;^macmon_about/$&#39;, &#39;macmonster.views.about&#39;),
        url(r&#39;^.*$&#39;,  &#39;macmonster.views.home&#39;),
    )





As it stands, the last entry sends all &quot;other&quot; traffic to the home page but I want to redirect via either an **HTTP 301** or **302**.

Django URL Redirect

After upgrading to Django 1.10, I get the error:

    TypeError: view must be a callable or a list/tuple in the case of include().

My urls.py is as follows:

    from django.conf.urls import include, url

    urlpatterns = [
        url(r&#39;^$&#39;, &#39;myapp.views.home&#39;),
        url(r&#39;^contact/$&#39;, &#39;myapp.views.contact&#39;),
        url(r&#39;^login/$&#39;, &#39;django.contrib.auth.views.login&#39;),
    ]

The full traceback is:

    Traceback (most recent call last):
      File &quot;/Users/alasdair/.virtualenvs/django110/lib/python2.7/site-packages/django/utils/autoreload.py&quot;, line 226, in wrapper
        fn(*args, **kwargs)
      File &quot;/Users/alasdair/.virtualenvs/django110/lib/python2.7/site-packages/django/core/management/commands/runserver.py&quot;, line 121, in inner_run
        self.check(display_num_errors=True)
      File &quot;/Users/alasdair/.virtualenvs/django110/lib/python2.7/site-packages/django/core/management/base.py&quot;, line 385, in check
        include_deployment_checks=include_deployment_checks,
      File &quot;/Users/alasdair/.virtualenvs/django110/lib/python2.7/site-packages/django/core/management/base.py&quot;, line 372, in _run_checks
        return checks.run_checks(**kwargs)
      File &quot;/Users/alasdair/.virtualenvs/django110/lib/python2.7/site-packages/django/core/checks/registry.py&quot;, line 81, in run_checks
        new_errors = check(app_configs=app_configs)
      File &quot;/Users/alasdair/.virtualenvs/django110/lib/python2.7/site-packages/django/core/checks/urls.py&quot;, line 14, in check_url_config
        return check_resolver(resolver)
      File &quot;/Users/alasdair/.virtualenvs/django110/lib/python2.7/site-packages/django/core/checks/urls.py&quot;, line 24, in check_resolver
        for pattern in resolver.url_patterns:
      File &quot;/Users/alasdair/.virtualenvs/django110/lib/python2.7/site-packages/django/utils/functional.py&quot;, line 35, in __get__
        res = instance.__dict__[self.name] = self.func(instance)
      File &quot;/Users/alasdair/.virtualenvs/django110/lib/python2.7/site-packages/django/urls/resolvers.py&quot;, line 310, in url_patterns
        patterns = getattr(self.urlconf_module, &quot;urlpatterns&quot;, self.urlconf_module)
      File &quot;/Users/alasdair/.virtualenvs/django110/lib/python2.7/site-packages/django/utils/functional.py&quot;, line 35, in __get__
        res = instance.__dict__[self.name] = self.func(instance)
      File &quot;/Users/alasdair/.virtualenvs/django110/lib/python2.7/site-packages/django/urls/resolvers.py&quot;, line 303, in urlconf_module
        return import_module(self.urlconf_name)
      File &quot;/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/importlib/__init__.py&quot;, line 37, in import_module
        __import__(name)
      File &quot;/Users/alasdair/dev/urlproject/urlproject/urls.py&quot;, line 28, in &lt;module&gt;
        url(r&#39;^$&#39;, &#39;myapp.views.home&#39;),
      File &quot;/Users/alasdair/.virtualenvs/django110/lib/python2.7/site-packages/django/conf/urls/__init__.py&quot;, line 85, in url
        raise TypeError(&#39;view must be a callable or a list/tuple in the case of include().&#39;)
    TypeError: view must be a callable or a list/tuple in the case of include().

Django URLs TypeError: view must be a callable or a list/tuple in the case of include()

I&#39;ve got two applications located on two separate computers. On computer A, in the `urls.py` file I have a line like the following: 

    (r&#39;^cast/$&#39;, &#39;mySite.simulate.views.cast&#39;)

And that url will work for both `mySite.com/cast/` and `mySite.com/cast`. But on computer B I have a similar url written out like:

    (r&#39;^login/$&#39;, &#39;mySite.myUser.views.login&#39;)

For some reason on computer B the `url mySite.com/login`/ will work but `mySite.com/login` will hang and won&#39;t direct back to `mySite.com/login/` like it will on computer A. Is there something I missed? Both `url.py` files look identical to me.

django urls without a trailing slash do not redirect

I have to build an url dynamically according to the current url. Using the `{% url %}` tag is the easiest way to do it, but I need the current url name to generate the new one dynamically. 

How can I get the url name attached to the urlconf that leads to the current view?

EDIT : I know I can manually handcraft the url using `get_absolute_url` but I&#39;d rather avoid it since it&#39;s part of a lecture and I would like to demonstrate only one way to build urls.

The students know how to use `{% url %}`. They are know facing a problem when they have to generate a more complete url based on the current one. The easiest way is to use `{% url %}` again, with some variations. Since we have named url, **we need to know how to get the name of the url that called the current view.**

EDIT 2: another use case is to display parts of the base template diff&#233;rently according to the base template. There are other ways to do it (using CSS and {% block %}, but sometime it just nice to be able to remove the &lt;a&gt;tag&lt;/a&gt; of the menu entry of base.html if the viewname match the link.

How to get the current url name using Django?

I have looked a lot on google for answers of how to use the &#39;url&#39; tag in templates only to find many responses saying &#39;You just insert it into your template and point it at the view you want the url for&#39;. Well no joy for me :( I have tried every permutation possible and have resorted to posting here as a last resort.

So here it is. My urls.py looks like this:

    from django.conf.urls.defaults import *
    from login.views import *
    from mainapp.views import *
    import settings
    
    # Uncomment the next two lines to enable the admin:
    from django.contrib import admin
    admin.autodiscover()
    
    urlpatterns = patterns(&#39;&#39;,
        # Example:
        # (r&#39;^weclaim/&#39;, include(&#39;weclaim.foo.urls&#39;)),
        (r&#39;^login/&#39;, login_view),
        (r&#39;^logout/&#39;, logout_view),
        (&#39;^$&#39;, main_view),
    
        # Uncomment the admin/doc line below and add &#39;django.contrib.admindocs&#39; 
        # to INSTALLED_APPS to enable admin documentation:
        # (r&#39;^admin/doc/&#39;, include(&#39;django.contrib.admindocs.urls&#39;)),
    
        # Uncomment the next line to enable the admin:
        (r&#39;^admin/&#39;, include(admin.site.urls)),
        #(r&#39;^static/(?P&lt;path&gt;.*)$&#39;, &#39;django.views.static.serve&#39;,{&#39;document_root&#39;: &#39;/home/arthur/Software/django/weclaim/templates/static&#39;}),
        (r&#39;^static/(?P&lt;path&gt;.*)$&#39;, &#39;django.views.static.serve&#39;,{&#39;document_root&#39;: settings.MEDIA_ROOT}),
    )

My &#39;views.py&#39; in my &#39;login&#39; directory looks like:

    from django.shortcuts import render_to_response, redirect
    from django.template import RequestContext
    from django.contrib import auth
    
    def login_view(request):
        if request.method == &#39;POST&#39;:
            uname = request.POST.get(&#39;username&#39;, &#39;&#39;)
            psword = request.POST.get(&#39;password&#39;, &#39;&#39;)
            user = auth.authenticate(username=uname, password=psword)
            # if the user logs in and is active
            if user is not None and user.is_active:
                auth.login(request, user)
                return render_to_response(&#39;main/main.html&#39;, {}, context_instance=RequestContext(request))
                #return redirect(main_view)
            else:
                return render_to_response(&#39;loginpage.html&#39;, {&#39;box_width&#39;: &#39;402&#39;, &#39;login_failed&#39;: &#39;1&#39;,}, context_instance=RequestContext(request))
        else:
            return render_to_response(&#39;loginpage.html&#39;, {&#39;box_width&#39;: &#39;400&#39;,}, context_instance=RequestContext(request))
    
    def logout_view(request):
        auth.logout(request)
        return render_to_response(&#39;loginpage.html&#39;, {&#39;box_width&#39;: &#39;402&#39;, &#39;logged_out&#39;: &#39;1&#39;,}, context_instance=RequestContext(request))

and finally the main.html to which the login_view points looks like:

    &lt;html&gt;
    &lt;body&gt;
    test! &lt;a href=&quot;{% url logout_view %}&quot;&gt;logout&lt;/a&gt;
    &lt;/body&gt;
    &lt;/html&gt;

So why do I get &#39;NoReverseMatch&#39; every time?

*(on a slightly different note I had to use &#39;context_instance=RequestContext(request)&#39; at the end of all my render-to-response&#39;s because otherwise it would not recognise {{ MEDIA_URL }} in my templates and I couldn&#39;t reference any css or js files. I&#39;m not to sure why this is. Doesn&#39;t seem right to me)*

Using {% url ??? %} in django templates

In a Django view you can access the `request.GET[&#39;variablename&#39;]`, so in your view you can do something like this:

    myvar = request.GET[&#39;myvar&#39;]

The actual `request.GET[&#39;myvar&#39;]` object type is:

    &lt;class &#39;django.http.QueryDict&#39;&gt;

Now, if you want to pass multiple variables with the same parameter name, i.e:

`http://example.com/blah/?myvar=123&amp;myvar=567`

You would like a python `list` returned for the parameter `myvar`, then do something like this:
 

    for var in request.GET[&#39;myvar&#39;]:
        print(var)

However, when you try that you only get the last value passed in the url i.e in the example above you will get **567**, and the result in the shell will be:

    5
    6
    7

However, when you do a print of `request.GET` it seems like it has a `list` i.e:

    &lt;QueryDict: {u&#39;myvar&#39;: [u&#39;123&#39;, u&#39;567&#39;]}&gt;


----------


Ok Update:
It&#39;s designed to return the last value, my use case is i need a list. 

from django docs:

&gt; QueryDict.__getitem__(key) 
&gt; Returns
&gt; the value for the given key. If the
&gt; key has more than one value,
&gt; __getitem__() returns the last value. Raises
&gt; django.utils.datastructures.MultiValueDictKeyError
&gt; if the key does not exist. (This is a
&gt; subclass of Python&#39;s standard
&gt; KeyError, so you can stick to catching
&gt; KeyError
&gt; 
&gt; QueryDict.getlist(key) Returns the
&gt; data with the requested key, as a
&gt; Python list. Returns an empty list if
&gt; the key doesn&#39;t exist. It&#39;s guaranteed
&gt; to return a list of some sort.

 Update: 
If anyone knows why django dev&#39;s have done this please let me know, seems counter-intuitive to show a list and it does not behave like one. Not very pythonic!



How to handle request.GET with multiple variables for the same parameter in Django

I&#39;m trying to create the back-end code for a music application on my website.

I have created the correct view in my views.py file (in the correct directory) as shown below:

    def detail(request, album_id):
        return HttpResponse(&quot;&lt;h1&gt;Details for Album ID:&quot; + str(album_id) + &quot;&lt;/h1&gt;&quot;)

However, when creating the URL or path for this (shown below)

    #/music/71/ (pk)
    path(r&#39;^(?P&lt;album_id&gt;[0-9])/$&#39;, views.detail, name=&#39;detail&#39;),

I am experiencing a warning on my terminal stating:
    
&gt; ```
&gt; ?: (2_0.W001) Your URL pattern &#39;^(?P&lt;album_id&gt;[0-9])/$&#39;
&gt; [name=&#39;detail&#39;] has a route that contains &#39;(?P&lt;&#39;, begins with a &#39;^&#39;,
&gt; or ends with a &#39;$&#39;. This was likely an oversight when migrating to
&gt; django.urls.path().
&gt; ```
And whenever the `/music/` (for which the path works) is followed by a number, such as `/music/1` (which is what I want to be able to do) the page cannot be found and the terminal gives the above warning.

Django 2.0 path error ?: (2_0.W001) has a route that contains &#39;(?P&lt;&#39;, begins with a &#39;^&#39;, or ends with a &#39;$&#39;

By default after login django redirects the user to an accounts/profile page or if you edit the LOGIN_REDIRECT_URL you can send the user to another page you specify in the settings.py.
 
This is great but I would like the user (after login) to be redirected to a custom page where the link to that page would look something like this: `mysite.com/username`. So the default accounts/profile or the LOGIN_REDIRECT_URL settings would not work in this case since both are somehow static. In my case the `username` section of the address changes for every user. 

Any ideas how I can make it so when the user is logged in would go to a custom user page that has user&#39;s name in the address like:  `mysite.com/username` ? Any input is truly appreciated. 
 



Django - after login, redirect user to his custom page --&gt; mysite.com/username

Is there a way to get the current page URL and all its parameters in a Django template? 

For example, a templatetag that would print a full URL like &lt;code&gt;/foo/bar?param=1&amp;baz=2&lt;/code&gt;




How to get URL of current page, including parameters, in a template?

After upgrading to Django 4.0, I get the following error when running `python manage.py runserver`

```
  ...
  File &quot;/path/to/myproject/myproject/urls.py&quot;, line 16, in &lt;module&gt;
    from django.conf.urls import url
ImportError: cannot import name &#39;url&#39; from &#39;django.conf.urls&#39; (/path/to/my/venv/lib/python3.9/site-packages/django/conf/urls/__init__.py)
```

My urls.py is as follows:

```
from django.conf.urls

from myapp.views import home

urlpatterns = [
    url(r&#39;^$&#39;, home, name=&quot;home&quot;),
    url(r&#39;^myapp/&#39;, include(&#39;myapp.urls&#39;),
]
```

ImportError: cannot import name &#39;url&#39; from &#39;django.conf.urls&#39; after upgrading to Django 4.0

I have a view that validates and saves a form.  After the form is saved, I&#39;d like redirect back to a list_object view with a success message &quot;form for customer xyz was successfully updated...&quot;  

HttpResponseRedirect doesn&#39;t seem like it would work, because it only has an argument for the url, no way to pass dictionary with it.

I&#39;ve tried modifying my wrapper for object_list to take a dict as a parameter that has the necessary context.  I the return a call to this wrapper from inside the view that saves the form.  However, when the page is rendered, the url is &#39;/customer_form/&#39; rather than &#39;/list_customers/&#39;.  I tried modifying the request object, before passing it to the object_list wrapper, but that did not work.

Thanks.

How do I redirect in Django with context?

I am pretty new to django but experienced in Python and java web programming with different frameworks.
I have made myself a nice little django app, but I cant seem to make it match www.mysite.com as opposed to www.mysite.com/myapp.

I have defined urls and views in my urls.conf which is currently not decoupled from the app (don&#39;t mind that). 

    urlpatterns = patterns(&#39;myapp.views&#39;,
      (r&#39;^myapp/$&#39;, &#39;index&#39;),
      (r&#39;^myapp/(?P&lt;some_id&gt;\d+)/global_stats/$&#39;, &#39;global_stats&#39;),
      (r&#39;^myapp/(?P&lt;some_id&gt;\d+)/player/(?P&lt;player_id&gt;\d+)/$&#39;, &#39;player_stats&#39;),
    )

All this works like a charm. If someone goes to www.mysite.com/myapp they will hit my index view which causes a http redirect to my &quot;correct&quot; default url.

So, how can I add a pattern that will do the same as (r&#39;^myapp/$&#39;, &#39;index&#39;) but without the /myapp--that is, www.mysite.com should suffice?

I would think this would be very basic stuff... I tried adding a line like:

    (r&#39;^$&#39;, &#39;index&#39;),

however this throws me in a loop... 

Hope you django gurus out there can clarify this for me!



Django URLS, how to map root to app?

Very basic question, but I&#39;m having trouble tracking down the answer on the web. I have a template, which I want to link to the django admin site (i.e. localhost:8000/admin). What is the code for this?

I&#39;m imagining something like

    &lt;a href=&quot;{% url admin.site.root %}&quot;&gt;link to admin panel&lt;/a&gt;

However, when I try the above snippet I get:

    Caught an exception while rendering:
      Reverse for &#39;project_name.django.contrib.admin.sites.root&#39; with
      arguments &#39;()&#39; and keyword arguments &#39;{}&#39; not found.

Help?



Linking to the django admin site

In rails, on can show the active routes with rake ([http://guides.rubyonrails.org/routing.html][1]):

    $ rake routes
              users GET  /users          {:controller=&gt;&quot;users&quot;, :action=&gt;&quot;index&quot;}
    formatted_users GET  /users.:format  {:controller=&gt;&quot;users&quot;, :action=&gt;&quot;index&quot;}
                    POST /users          {:controller=&gt;&quot;users&quot;, :action=&gt;&quot;create&quot;}
                    POST /users.:format  {:controller=&gt;&quot;users&quot;, :action=&gt;&quot;create&quot;}

Is there a similar tool/command for django showing the e.g. the URL pattern, the name of the pattern (if any) and the associated function in the views?

  [1]: http://guides.rubyonrails.org/routing.html

Is there something similar to &#39;rake routes&#39; in django?

Is possible to add GET variables in a redirect ? (Without having to modifiy my urls.py) 

If I do `redirect(&#39;url-name&#39;, x)`

I get `HttpResponseRedirect(&#39;/my_long_url/%s/&#39;, x)`

I don&#39;t have complains using `HttpResponseRedirect(&#39;/my_long_url/%s/?q=something&#39;, x)` instead, but just wondering...

add request.GET variable using django.shortcuts.redirect

I have achieved 100% test coverage in my application everywhere except my `urls.py`. Do you have any recommendations for how I could write *meaningful* unit tests for my URLs?

FWIW This question has arisen as I am experimenting with Test-Driven Development and want failing tests before I write code to fix them.

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