Question

I have a Model class.

class Search(forms.Form):
    query=forms.CharField()

And, a view function.

def search(request):
  c = {}
  c.update(csrf(request))
  if request.method == 'POST': 
    form = Search(request.POST) 
      if form.is_valid():
        search_query=form.cleaned_data['query']
            return HttpResponse("your query: %s" %search_query  , c) 
      else:
        return HttpResponse(form , c)
  else:
    return render_to_response('polls/search.html', c)

And, here is my search.html:

 <form action="/polls/search" method="post">{% csrf_token %}
  <p><label for="query_label">query:</label>
    <input type="text" name="query_txt" id="query_txt_id" /></p>
  <input type="submit" value="Submit" />
 </form>

After giving some characters as input, it always show me the following as plain text(Not as error).

“A server error occurred. Please contact the administrator.”

Answer 1

The id of your query field is incorrect. Django expects it to be id_query.

You don’t need to hardcode your form inputs. If you include {{ form.as_p }} in your template, Django will render the form correctly.

<form action="/polls/search" method="post">{% csrf_token %}
  {{ form.as_p }}
  <input type="submit" value="Submit" />
</form>

If you really want to hardcode the form in the template, start with the working html that Django produces, and customize it from there. Note that the forms in your template and the snippet below do not display errors. See the docs on customizing the form template for more details.

<form action="/polls/search" method="post">{% csrf_token %}
  <p><label for="id_query">Query:</label> <input type="text" name="query" id="id_query" /></p>
  <input type="submit" value="Submit" />
</form>