django: taking input and showing output in the same page - django

Question

I am quite new to django and struggling to do something very simple.
I have a ModelForm for the following model:
class Queries(models.Model):
user_id=models.CharField(max_length=200)
query=models.CharField(max_length=200)
And I am showing the user a simple form that will help in doing the following:
user will ask a question
The question will be processed(a database query will be generated
based on the question)
Then the query result should be shown just beneath the form in the
same page.
This is how my views.py looks like:
from django.http import HttpResponse
from django.shortcuts import get_object_or_404, render
from basicapp.models import QueryForm
def index(request):
form=MyForm()
real_form=form.getForm(request)
response=form.response
return render(request,'basicapp/index.html',{
'form': real_form,
'response':response,
})
class MyForm:
response=''
def getForm(self,request):
form = QueryForm(request.POST)
if form.is_valid():
response=form.cleaned_data['query']
form.save()
return form
For now I am trying simple stuffs,I am taking the value in query field of the form and trying to send it back to the page;so far I am failed.
This is index.html:
<form action=" " method="post">{% csrf_token %}
{{ form }}
<p>{{response}}</p>
<input type="submit" value="Submit" />
</form>
If I could do this,I think the query stuffs wont be that tough.The form is working fine,the datas are getting saved in database. Only the response string from views.py could not be retrieved inside index.html after form submission. Can you please help?
EDIT:
Tried following in index.html based on Hoff's answer:
<form id="myForm" action=" " method="get">{% csrf_token %}
{{ form }}
<input type="submit" value="Submit" />
</form>
<div id="response">
</div>
<script language="JavaScript">
$(document).ready(function() {
$("#myForm").submit(function() { // catch the form's submit event
$.ajax({ // create an AJAX call...
data: $(this).serialize(), // get the form data
type: $(this).attr('GET'),
success: function(response) { // on success..
$("#response").html(response); // update the DIV
}
});
return false;
});
});
</script>
Still no luck :(

Answer 1

views.py
def index(request):
questions=None
if request.GET.get('search'):
search = request.GET.get('search')
questions = Queries.objects.filter(query__icontains=search)
name = request.GET.get('name')
query = Queries.object.create(query=search, user_id=name)
query.save()
return render(request, 'basicapp/index.html',{
'questions': questions,
})
html
<form method="GET">
Question: <input type="text" name="search"><br/>
Name: <input type="text" name="name"><br/>
<input type="submit" value="Submit" />
</form><br/><br/>
{% for question in questions %}
<p>{{question}}</p>
{% endfor %}

Answer 2

What you need is an asynchronous post (ajax), which is easy with jQuery, see this answer for a complete solution: How to POST a django form with AJAX & jQuery

Answer 3

<input type="text" name="query" />
<input type="submit" name="submit" value="Submit" />
you can check if the form was submitted or not (i.e if it's a post request or not):
if 'submit' in request.POST: #you could use 'query' instead of 'submit' too
# do post related task
# add context variables to render post output
# add another context variable to indicate if it's a post
# Example:
context.update({'post_output': request.POST.get('query','')})
...
return render(request, 'index.html', context)
Then in the template, check if context variable post_output exists, if it does show the output:
{% if post_output %}
Output: {{ post_output }}
{% endif %}
In short, the logic is:
Check if a relevant request.POST dict key exists or not in your view.
If the key exists, then it's a post request; add post related context variables and do post related tasks.
Check if any post related context variable is available in the template and if it does, show post related output.
If you don't want to show the output when the page is simply refreshed after a post, pass the request object to the template and do a check like this:
{% if request.POST.submit and post_output %}

Answer 4

Following Hoff's answer...
Add URL attribute to ajax call:
$(document).ready(function() {
$("#myForm").submit(function() { // catch the form's submit event
$.ajax({ // create an AJAX call...
data: $(this).serialize(), // get the form data
type: $(this).attr('GET'),
url: '/URL-to-ajax-view/',
success: function(response) { // on success..
$("#response").html(response); // update the DIV
}
});
return false;
});
});
Some ajax handler in views.py:
# /URL-to-ajax-view/
def ajax_get_response(request):
if request.method == "GET" and request.is_ajax:
form = QueryForm(request.POST or None)
if form.is_valid():
form.save()
return HttpResponse(form.response)
raise Http404
Tried something like that?