[Solved]-Django: taking input and showing output in the same page

by

12👍

✅

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 %}
👤catherine

3👍

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

👤Hoff

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:

  1. Check if a relevant request.POST dict key exists or not in your view.
  2. If the key exists, then it’s a post request; add post related context variables and do post related tasks.
  3. 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 %}
👤Jahid

1👍

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?

👤Ogre

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