The other answers require you to pass an additional context variable. But as long as you can access the request object, there is no need:

    {% extends request.is_ajax|yesno:&quot;app/base_ajax.html,app/base.html&quot; %}

I found this to be much more convenient.

Use a variable.

    {% extends base_template %}

and in your view, set it to &quot;base.html&quot; in your view, or a new &quot;ajax.html&quot; file which just provides the block and nothing else.


    {% extends override_base|default:&#39;base.html&#39; %}

P.s. I know this is an old question, but I found it when searching for an answer.  Maybe it&#39;ll help someone else with the same problem.

You can use {% extends variable %}

Pass a variable base template name in when you create the context in the view.

http://docs.djangoproject.com/en/dev/ref/templates/builtins/?from=olddocs#extends

The image with invalid source displays an alternate text in Firefox but not in chrome unless the width of an image is adjusted.

      &lt;img height=&quot;90&quot; width=&quot;90&quot;
		src=&quot;http://www.google.com/intl/en_ALL/images/logos/images_logo_lg.gif&quot;
		alt=&quot;Image Not Found&quot;/&gt;


How to display the alt text for an image?


How to display alt text for an image in chrome

I am launching a beta site with a select group of users. I want to disable registration in the production environment only, and only for a short period of time (i.e. I don&#39;t want to nuke my registration altogether). I know I can simply hide the &quot;sign up&quot; link, but I suspect that hackers smarter than I can still use the RESTful routes to accomplish registrations. What&#39;s the best way to disable registration so my test/development environments still work, but production is affected? Thanks for any pointers.

I&#39;ve tried pointing named scopes in such a way that &quot;sign_up&quot; goes to &quot;sign_in&quot;, but it didn&#39;t work. Here&#39;s what I&#39;ve tried:

    devise_scope :user do
        get &quot;users/sign_in&quot;, :to =&gt; &quot;devise/sessions#new&quot;, :as =&gt; :sign_in
        get &quot;users/sign_up&quot;, :to =&gt; &quot;devise/sessions#new&quot;, :as =&gt; :sign_up
    end

Ideally, we&#39;d send the user to a &quot;pages#registration_disabled&quot; page or something like that. I just wanted to get something working I can play around with.

EDIT:
I&#39;ve changed the model as requested, then added the following to /spec/user_spec.rb

    describe &quot;validations&quot; do
        it &quot;should fail registration if in production mode&quot; do
          ENV[&#39;RAILS_ENV&#39;] = &quot;production&quot;
          @user = Factory(:user).should_not be_valid
        end
    end

it is passing as &quot;true&quot; rather than false. Is there a way to mock up the production environment? I&#39;m just spit-balling this one. 

Thanks!

disabling Devise registration for production environment only

I would like to share a template between AJAX and regualr HTTP calls, the only difference is that one template needs to be served with the base.html html, the other one without.

Any idea?

Any way to make {% extends &#39;...&#39; %} conditional? - Django

<p>I would like to share a template between AJAX and regualr HTTP calls, the only difference is that one template needs to be served with the base.html html, the other one without.</p>
<p>Any idea?</p>


If I want the number of items in an iterable without caring about the elements themselves, what would be the pythonic way to get that? Right now, I would define

    def ilen(it):
        return sum(itertools.imap(lambda _: 1, it))    # or just map in Python 3

but I understand `lambda` is close to being considered harmful, and `lambda _: 1` certainly isn&#39;t pretty.

(The use case of this is counting the number of lines in a text file matching a regex, i.e. `grep -c`.)

What&#39;s the shortest way to count the number of items in a generator/iterator?

Is it legitimate to delete items from a dictionary in Python while iterating over it?

For example:

    for k, v in mydict.iteritems():
       if k == val:
         del mydict[k]

The idea is to remove elements that don&#39;t meet a certain condition from the dictionary, instead of creating a new dictionary that&#39;s a subset of the one being iterated over.

Is this a good solution? Are there more elegant/efficient ways?

How to delete items from a dictionary while iterating over it?

How do you convert a string into a list?

Say the string is like `text = &quot;a,b,c&quot;`. After the conversion, `text == [&#39;a&#39;, &#39;b&#39;, &#39;c&#39;]` and hopefully `text[0] == &#39;a&#39;`, `text[1] == &#39;b&#39;`?

How to convert a string with comma-delimited items to a list in Python?

Is there a way in Python `unittest` to set the order in which test cases are run?

In my current `TestCase` class, some testcases have side effects that set conditions for the others to run properly. Now I realize the proper way to do this is to use `setUp()` to do all setup related things, but I would like to implement a design where each successive test builds slightly more state that the next can use. I find this much more elegant.

    class MyTest(TestCase):

      def test_setup(self):
        # Do something

      def test_thing(self):
        # Do something that depends on test_setup()

Ideally, I would like the tests to be run in the order they appear in the class. It appears that they run in alphabetical order.





Python unittest.TestCase execution order

I am parsing an XSL file using xlrd. Most of the things are working fine. I have a dictionary where keys are strings and values are lists of strings. All the keys and values are Unicode. I can print most of the keys and values using `str()` method. But some values have the Unicode character `\u2013` for which I get the above error. 

I suspect that this is happening because this is Unicode embedded in Unicode and the Python interpreter cannot decode it. So how can I get rid of this error? 


UnicodeEncodeError: &#39;ascii&#39; codec can&#39;t encode character u&#39;\u2013&#39; in position 3 2: ordinal not in range(128)

I&#39;ve got a problem sending a file to a serverside PHP-script using jQuery&#39;s ajax-function.
It&#39;s possible to get the File-List with `$(&#39;#fileinput&#39;).attr(&#39;files&#39;)` but how is it possible to send this Data to the server? The resulting array (`$_POST`) on the serverside php-script is 0 (`NULL`) when using the file-input.

I know it is possible (though I didn&#39;t find any jQuery solutions until now, only Prototye code ([http://webreflection.blogspot.com/2009/03/safari-4-multiple-upload-with-progress.html][1])).

This seems to be relatively new, so please do not mention file upload would be impossible via XHR/Ajax, because it&#39;s definitely working.

I need the functionality in Safari 5, FF and Chrome would be nice but are not essential.

My code for now is:

    $.ajax({
        url: &#39;php/upload.php&#39;,
        data: $(&#39;#file&#39;).attr(&#39;files&#39;),
        cache: false,
        contentType: &#39;multipart/form-data&#39;,
        processData: false,
        type: &#39;POST&#39;,
        success: function(data){
            alert(data);
        }
    });


  [1]: http://webreflection.blogspot.com/2009/03/safari-4-multiple-upload-with-progress.html

Sending multipart/formdata with jQuery.ajax

The following `attempt_login` method is called using Ajax after a login form is submitted.

    class AccessController &lt; ApplicationController
      [...]
      def attempt_login
        authorized_user = User.authenticate(params[:username], params[:password])
        
        if authorized_user
          session[:user_id] = authorized_user.id
          session[:username] = authorized_user.username
          flash[:notice] = &quot;Hello #{authorized_user.name}.&quot;
          redirect_to(:controller =&gt; &#39;jobs&#39;, :action =&gt; &#39;index&#39;)
        else
          [...]
        end
      end
    end

The problem is that `redirect_to` doesn&#39;t work.

How would you solve this ?

Rails 3: How to &quot;redirect_to&quot; in Ajax call?

Is there a way that I can see the URL that was requested when I do an Ajax request with jQuery?

e.g.,

    var some_data_object = { ...all sorts of junk... }
    $.get(&#39;/someurl.php&#39;,some_data_object, function(data, textStatus, jqXHR) {
       var real_url = ? # &lt;-- How do I get this
    })

How can I access the URL that jQuery actually used to make the request? Perhaps some method/property of `jqHXR`? I couldn&#39;t find it in the documentation.

Thanks.

Access the URL of an jQuery Ajax Request in the Callback Function

I am trying to create a basic authentication through the browser, but I can&#39;t really get there.

If this script won&#39;t be here the browser authentication will take over, but I want to tell the browser that the user is about to make the authentication.

The address should be something like:

    http://username:password@server.in.local/

I have a form:

    &lt;form name=&quot;cookieform&quot; id=&quot;login&quot; method=&quot;post&quot;&gt;
          &lt;input type=&quot;text&quot; name=&quot;username&quot; id=&quot;username&quot; class=&quot;text&quot;/&gt;
          &lt;input type=&quot;password&quot; name=&quot;password&quot; id=&quot;password&quot; class=&quot;text&quot;/&gt;
          &lt;input type=&quot;submit&quot; name=&quot;sub&quot; value=&quot;Submit&quot; class=&quot;page&quot;/&gt;
    &lt;/form&gt;

And a script:

    var username = $(&quot;input#username&quot;).val();
    var password = $(&quot;input#password&quot;).val();

    function make_base_auth(user, password) {
      var tok = user + &#39;:&#39; + password;
      var hash = Base64.encode(tok);
      return &quot;Basic &quot; + hash;
    }
    $.ajax
      ({
        type: &quot;GET&quot;,
        url: &quot;index1.php&quot;,
        dataType: &#39;json&#39;,
        async: false,
        data: &#39;{&quot;username&quot;: &quot;&#39; + username + &#39;&quot;, &quot;password&quot; : &quot;&#39; + password + &#39;&quot;}&#39;,
        success: function (){
        alert(&#39;Thanks for your comment!&#39;);
        }
    });


Use basic authentication with jQuery and Ajax

I have a text field which is showing a value. What I want is when the user writes a new value in that text field and presses enter, an ajax function is triggered to do a pagination operation. I have a text field like this:

    &lt;input type=&quot;text&quot; id=&quot;page&quot; name=&quot;page&quot; value=&quot;&lt;?php echo($this-&gt;pn);?&gt; /&gt;

And when a user writes any new value and presses Enter, I want the following ajax function triggered:
&lt;pre&gt;
update_ajax2({rpp:&amp;lt;?php echo($this-&gt;rpp);?&gt;,pn:document.page.paged.value,filter:&#39;&amp;lt;?php echo($this-&gt;filter);?&gt;&#39;,orderby:&#39;&amp;lt;?php echo($this-&gt;orderby);?&gt;&#39;});
&lt;/pre&gt;

I tried using the keypress event to detect `if(e.which===13)` but this doesn&#39;t solve the problem. Can anyone guide me?

Detect Enter key is pressed with jquery

I find the the Django Admin&#39;s default `models.ManyToManyField` widget to be cumbersome to use. It&#39;s the [HTML select](http://www.htmlcodetutorial.com/forms/_SELECT_MULTIPLE.html) element and if you have a lot of Objects of the &quot;other&quot; model then it&#39;s quite impractical to actually find the &quot;other&quot; Objects you want to associate with &quot;this&quot; Object. And if you have ***a lot*** of objects of the &quot;other&quot; model it seems to even slows down the rendering of the Admin page.

I&#39;m aware that I can build my own custom admin widget and apply it to my `ManyToManyFields` as I see fit, but are there any pre-built ones out there that I might use instead? In my dreams, I picture an auto-completing text input HTML widget. Is this even practical/possible to do in the Django admin framework?

Thanks.

A Better Django Admin ManyToMany Field Widget

I&#39;m new to Python and Django.

I&#39;m configuring a Django project using a PostgreSQL database engine backend, But I&#39;m getting errors on each database operation. For example when I run `manage.py syncdb`, I&#39;m getting:

	C:\xampp\htdocs\djangodir&gt;python manage.py syncdb
	Traceback (most recent call last):
	  File &quot;manage.py&quot;, line 11, in &lt;module&gt;
		execute_manager(settings)
	  File &quot;C:\Python27\lib\site-packages\django\core\management\__init__.py&quot;, line
	438, in execute_manager
		utility.execute()
	  File &quot;C:\Python27\lib\site-packages\django\core\management\__init__.py&quot;, line
	379, in execute
		self.fetch_command(subcommand).run_from_argv(self.argv)
	  File &quot;C:\Python27\lib\site-packages\django\core\management\__init__.py&quot;, line
	261, in fetch_command
		klass = load_command_class(app_name, subcommand)
	  File &quot;C:\Python27\lib\site-packages\django\core\management\__init__.py&quot;, line
	67, in load_command_class
		module = import_module(&#39;%s.management.commands.%s&#39; % (app_name, name))
	  File &quot;C:\Python27\lib\site-packages\django\utils\importlib.py&quot;, line 35, in im
	port_module
		__import__(name)
	  File &quot;C:\Python27\lib\site-packages\django\core\management\commands\syncdb.py&quot;
	, line 7, in &lt;module&gt;
		from django.core.management.sql import custom_sql_for_model, emit_post_sync_
	signal
	  File &quot;C:\Python27\lib\site-packages\django\core\management\sql.py&quot;, line 6, in
	 &lt;module&gt;
		from django.db import models
	  File &quot;C:\Python27\lib\site-packages\django\db\__init__.py&quot;, line 77, in &lt;modul
	e&gt;
		connection = connections[DEFAULT_DB_ALIAS]
	  File &quot;C:\Python27\lib\site-packages\django\db\utils.py&quot;, line 92, in __getitem
	__
		backend = load_backend(db[&#39;ENGINE&#39;])
	  File &quot;C:\Python27\lib\site-packages\django\db\utils.py&quot;, line 33, in load_back
	end
		return import_module(&#39;.base&#39;, backend_name)
	  File &quot;C:\Python27\lib\site-packages\django\utils\importlib.py&quot;, line 35, in im
	port_module
		__import__(name)
	  File &quot;C:\Python27\lib\site-packages\django\db\backends\postgresql\base.py&quot;, li
	ne 23, in &lt;module&gt;
		raise ImproperlyConfigured(&quot;Error loading psycopg module: %s&quot; % e)
	django.core.exceptions.ImproperlyConfigured: Error loading psycopg module: No mo
	dule named psycopg

Can someone give me a clue on what is going on?

How to set up a PostgreSQL database in Django

    {% for each in AnswerQuery %}
        &lt;form action={{address}}&gt;
            &lt;span&gt;{{each.answer}}&lt;/span&gt;&lt;input type=&#39;radio&#39;&gt;
            &lt;span&gt;Votes:{{each.answercount}}&lt;/span&gt;
            &lt;br&gt;
        &lt;/form&gt;
    {% endfor %}

This is a part my [django][1] template, what it supposed to do is to print out several radio buttons, corresponding to the answers assigned to the buttons. But I don&#39;t know why I can check multiple radio buttons, which messed me up. It is supposed to only let me check on one radio button and I had that somehow but I lost it. Any help? Thank you. 


  [1]: https://www.djangoproject.com/

How to allow only one radio button to be checked?

If, for a field that you want to filter by, you have more than ~10 values, the filtering sidebar starts to be ugly and harder to use.

I&#39;m looking for a solution to replace the `&lt;li&gt;` with a dropdown selection (combobox) or something similar that will solve the same problem.

How to change the Django admin filter to use a dropdown instead of list?

I have the following Django model:

    class Make:
       name = models.CharField(max_length=200)
    
    class MakeContent:
       make = models.ForeignKey(Make)
       published = models.BooleanField()

I&#39;d like to know if it&#39;s possible (without writing SQL directly) for me to generate a queryset that contains all `Make`s and each one&#39;s related `MakeContent`s where `published = True`.


Django Queryset with filtering on reverse foreign key

I have implemented a password recovery functionality in django. With my method, the new password will be sent to the email id entered. It works fine when given the correct email (e-mail id which exists in the database). But when given an email id which is not in the database, it gives the error:  
    `&#39;DoesNotExist at /forgotPassword/  UniversityDetails matching query does not exist.&#39;`

How can I resolve this issue?

**forgotPassword.html()**

    def forgotPassword(request):
        if request.POST:
            email=request.POST.get(&quot;email&quot;)
            user = UniversityDetails.objects.get(email=email)
            if(not user):
                print &quot;No user&quot;
                return render_to_response(&quot;forgotPassword.html&quot;)
            else:
                newPassword = user.password
                send_mail(&#39;Password Recovery&#39;, &#39;The password for your site is &#39;+ newPassword, &#39;rv_nair@gmail.com&#39;,
        [&#39;rv_ks@gmail.com&#39;], fail_silently=False)   
                return render_to_response(&quot;passwordRecovery.html&quot;)
        return render_to_response(&#39;forgotPassword.html&#39;)

**html**

    &lt;form name=&quot;forgotPassword&quot; method=&quot;POST&quot; id=&quot;myFormid&quot; action=&quot;http://10.1.0.90:8080/forgotPassword/&quot;&gt;
    &lt;div style=&quot;float:center;width:100%;color:#0000A0&quot;&gt;
     Enter your E-mail ID&lt;/label&gt;&lt;br/&gt; &lt;input type=&quot;text&quot; name=&quot;email&quot; size=&quot;25&quot; /&gt; 
     &lt;input type=&quot;submit&quot; value=&quot;Submit&quot; /&gt;
     &lt;/div&gt; 
    
    &lt;/form &gt;

matching query does not exist Error in Django

I want to do this:

    100 - {{ object.article.rating_score }} 

So for example, the output would be `20` if `{{ object.article.rating_score }}` equaled `80`.

How do I do this at the template level? I don&#39;t have access to the Python code.

How to do math in a Django template?

I have _users_, _videos_, _topics_, _criterias_ and _ratings_

- A _video_ has a _topic_
- A _topic_ has _criterias_
- A _user_ can create a _video_ for a given _topic_
- A _user_ can rate a _video_ on each _criterias_ given for the concerned _topic_.

You can see my original post https://stackoverflow.com/questions/6388259/django-rating-model-example-detailview-template to get details on the model used

I have extended a `DetailView` template based on the _video_ model to put the list of _ratings_ for the selected _video_ for a given _user_ as extra context.

    class VideoFileDetailView(DetailView):
      model = VideoFile

      def get_context_data(self, **kwargs):
        context = super(VideoFileDetailView, self).get_context_data(**kwargs)
        context[&#39;rates&#39;] = VideoRate.objects.filter(video=self.object, user=self.request.user)
        return context

In the template pointed by the `DetailView`, I&#39;d like to list the _criterias_ of the _video_, and for each _criteria_ display the current **rating value** form the _user_.


    &lt;div id=&quot;rating&quot;&gt;
      &lt;ul&gt;
    {% for crit in videofile.topic.crits.all %}
        &lt;li&gt;
      {% for rate in rates %}
        {% if rate.crit.id == crit.id %}
          {{ rate.rate }}
        {% endif %}
      {% endfor %}
          &lt;div class=&quot;rateit&quot;
            data-rateit-value=&quot;{# The rating value #}&quot;
            data-rateit-ispreset=&quot;true&quot;
            crit-id=&quot;{{ crit.id }}&quot;&gt;&lt;/div&gt;
          {{ crit }}
        &lt;/li&gt;
    {% endfor %}
      &lt;/ul&gt;
    &lt;/div&gt;

([rateit][1] is a jquery plugin that I use to draw pretty stars rating controls)

Actually I get my rating values here within the 2nd `for` but I&#39;m sure there is a better way to do that. In fact, I&#39;m still not sure about my [model][2] correctness.

Finally I&#39;d like to replace `{# The rating value #}` by the rating value from rate for the current crit (in the loop). How can I do that ?

  [1]: http://www.radioactivethinking.com/rateit/example/example.htm
  [2]: https://stackoverflow.com/questions/6388259/django-rating-model-example-detailview-template

Django - Simple custom template tag example

Is it possible to sort a set of related items in a DJango template?

That is: this code (with HTML tags omitted for clarity):

    {% for event in eventsCollection %}
       {{ event.location }}
       {% for attendee in event.attendee_set.all %}
         {{ attendee.first_name }} {{ attendee.last_name }}
       {% endfor %}
     {% endfor %}

displays ***almost*** exactly want I want.  The only thing I want to change is I the list of attendees to be sorted by last name.  I&#39;ve tried saying something like this:

    {% for event in events %}
       {{ event.location }}
       {% for attendee in event.attendee_set.order_by__last_name %}
         {{ attendee.first_name }} {{ attendee.last_name }}
       {% endfor %}
     {% endfor %}

Alas, the above syntax doesn&#39;t work (it produces an empty list) and neither does any other variation I have thought of (lot&#39;s of syntax errors reported, but no joy).   

I could, of course, produce some kind of array of sorted attendee lists in my view, but that is an ugly and fragile (and did I mention ugly) solution.

Needless to say, but I&#39;ll say it anyway, I have perused the on-line docs and searched Stack Overflow and the archives of django-user without finding anything helpful (ah, if only a query set were a dictionary dictsort would do the job, but it&#39;s not and it doesn&#39;t)

==============================================

Edited to add additional thoughts
after accepting Tawmas&#39;s answer.

-----------------------------------------------
Tawmas addressed the issue exactly as I presented it -- although the solution was not what I expected.  As a result I learned a useful technique that can be used in other situations as well. 

Tom&#39;s answer proposed an approach I had already mentioned in my OP and tentatively rejected as being &quot;ugly&quot;.  

The &quot;ugly&quot; was a gut reaction, and I wanted to clarify what was wrong with it.  In doing so I realized that the reason it was an ugly approach was because I was hung up on the idea of passing a query set to the template to be rendered.   If I relax that requirement, there is an un-ugly approach that should work.  

I haven&#39;t tried this yet, but suppose that rather than passing the queryset, the view code iterated through the query set producing a list of Events, then decorated each Event with a query set for the corresponding attendees which _WAS_ sorted (or filtered, or whatever) in the desired way.   Something like so:
 
    eventCollection = []   
    events = Event.object.[filtered and sorted to taste]
    for event in events:
       event.attendee_list = event.attendee_set.[filtered and sorted to taste]
       eventCollection.append(event)

Now the template becomes:

    {% for event in events %}
       {{ event.location }}
       {% for attendee in event.attendee_list %}
         {{ attendee.first_name }} {{ attendee.last_name }}
       {% endfor %}
     {% endfor %}


The downside is the view has to &quot;actualize&quot; all of the events at once which could be a problem if there were large numbers of events.  Of course one could add pagination, but that complicates the view considerably.  

The upside is the &quot;prepare the data to be displayed&quot; code is in the view where it belongs letting the template focus on formatting the data provided by the view for display. This is right and proper.
       
So my plan is to use Tawmas&#39; technique for large tables and the above technique for small
tables, with the definition of large and small left to the reader (grin.)
        


Sorting related items in a Django template

I have this code

    {% for o in some_list %}

Now I want to do some stuff if I am on an even line. How can I do that?

Content Type	Original Author	Original Content on Stackoverflow
Question	RadiantHex	View Question on Stackoverflow
Solution 1 - Python	Cruel	View Answer on Stackoverflow
Solution 2 - Python	Daniel Roseman	View Answer on Stackoverflow
Solution 3 - Python	Brigand	View Answer on Stackoverflow
Solution 4 - Python	Tom Gruner	View Answer on Stackoverflow

Any way to make {% extends '...' %} conditional? - Django

Python Problem Overview

Python Solutions

Solution 1 - Python

Solution 2 - Python

Solution 3 - Python

Solution 4 - Python

disabling Devise registration for production environment only

How to display alt text for an image in chrome

Attributions