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Update DOM without reloading the page in Django

Tags:

python

django

I have two picklists in DOM, first is populated with initial call to a view (i.e. loading the page). Second picklist's content depends on what user selects in the first one.

Let's say the picklists are:

  1. list_of_events
  2. list_of_rooms

Every time the state of the first picklist changes (user selects a different event) I need to update the content of the second picklist (available rooms) without refreshing the page. Is this possible?

My template looks like this:

{% block content %}

{% if list_of_events %}
    <form>
        <select>
            {% for event in list_of_events %}
                <option value="name">{{ event.title }}</option>
            {% endfor %}
        </select>
    </form>

{% else %}
    <p>No events available now.</p>
{% endif %}

For now the second picklist list_of_rooms should only be visible if user have selected anything from the list_of_events picklist. first_selection argument is passed to template if user selects a value in list_of_events.

{% if first_selection %}
    <form>
        <select>
            {% for room in list_of_rooms %}
                <option value="name">{{ room.id }}</option>
            {% endfor %}
        </select>
    </form>
{% endif %}

<input type="submit" value="Submit">

{% endblock %}

My views.py function looks like this:

def events(request):
    list_of_events = Event.objects.all()
    return render(
        request, 
        'web_service/schedule.html', {'list_of_events': list_of_events}
    )
like image 567
Stan Redoute Avatar asked Aug 27 '17 16:08

Stan Redoute


2 Answers

In Django, at least now, there's no direct way to dynamically call python method from html template without refreshing the page.

To call python method and see it's effect in template without refreshing the page you'll need a bit of JS, dedicated url pattern and a view. It’s not as simple as calling instance method, but not as hard as it may seem.

The example below is a demonstration how to respond to a button click, send some data to a view and return the result to template without refreshing DOM.


The only way to call python code from template is relate to in the same way as we relate to url, this means we have to create new url pattern. Then call necessary view and return response to template as JsonResponse.

Note: make sure to import jquery at the inside bottom of your <body> tag.

First of all we need to create responder which will handle button click and create AJAX request to url connected to view. AJAX request passes input as parameter to our url pattern, meaning this parameter will be passed to django view. If something returns from our call to view then data is being unpacked in success closure.

Let’s say our template looks like this:

<input type="text" id="user-input" autofocus=""><br>
<button type="button" id="sender">Send data</button><br>
<p id="p-text">foo bar</p>

Script handling clicks and request looks like this:

<script>

$("#sender").click(function () {
    var input = $('#user-input').val();

    $.ajax({
        url: '{% url 'get_response' %}',
        data: {
          'inputValue': input
        },
        dataType: 'json',
        success: function (data) {
          document.getElementById('p-text').innerHTML = data["response"];
        }
      });
    });

</script>

New pattern is needed in urls.py:

urlpatterns = [
    ...
    url(r'^ajax/get_response/$', views.answer_me, name='get_response')
    ...
]

Note: the ajax/ part of url and path of url pattern itself has no influence on how this call is handled. It can be anything you want, for example: ^foo/bar/$.

Last step is adding responding Django view. This simple example returns the input with some additional text, but generally this is the place where you can call other python methods and do whatever you want:

def answer_me(request):
    user_input = request.GET.get('inputValue')
    data = {'response': f'You typed: {user_input}'}
    return JsonResponse(data)
like image 144
Stan Redoute Avatar answered Nov 10 '22 13:11

Stan Redoute


I had a similar issue to OP (the accepted answer is the closest I can find to what I came up), and this was the top hit to my google search, so I'll share what I came up with.

The main difference is I set the dataType to 'html' and just appended the rendered template directly to the element. (if you didnt change the dataType you'd end up with exceptions--wouldn't work in the manner I've implemented)

//parent_template.html
{% load static %}
<script src="{% static "jquery.js" %}"></script>

{% block first %}
<div class="row">
    {% for x in my_list %}
    <button class="some-filter" id="{{x}}"></button>
    {% endfor %}
</div>
{% endblock first %}

{% block second %}
<div class="row" id="some_id"></div>
{% endblock second %}

<script>
     $(".some-filter").on({
       click: e=> {
           var name = $( e.target )[0].id;
           $.ajax({
             url: '/ajax/info_getter/',
             data: {
               'name': name,
             },
             dataType: 'html',
             success: function (data) {
               if (data) {
                 $("#some_id").html(data);
                 $(".some-filter").removeClass('selected');
                 $( e.target ).addClass('selected');
               }
             }
           });
       }
   })
</script>



// child_template.html
// my big ass template containing lots of image files
{% load static %}
...other stuff

// urls.py
from django.urls import path
from . import views
urlpatterns = [
    ...
    path('ajax/info_getter/', views.info_getter, name='info_getter'),
    ...
]



// views.py
from django.shortcuts import render
...
def info_getter(request):
    name = request.GET.get('name', None)
    if name:
        context["name"] = name

    response = render(request, "child_template.html", context=context)
    return response

Note: I'm not sure if this is considered best practice

Note: only tested on Django 2.2 and Python 3.6.x

like image 45
kt-0 Avatar answered Nov 10 '22 13:11

kt-0