You could cache the output, some thing like:
def generate_questionnaire_view():
if is_cached_output_available():
return get_cached_oputput()
else:
output_html = render_to_string(your parameters)
add_to_cache(output_html)
return output_html
The is_cached_output_available() function will check if we have some output
cached, and nothing has changed since we cached the output.
Regards,
Mayuresh
http://twitter.com/geeroo
http://twitter.com/django_updates
On Monday, December 27, 2010 8:08:40 PM UTC+5:30, Guax3 wrote:
>
> I have a template in where I load a set of questions for a a
> questionnaire, and each question has a set of options all this
> information is retrieved from a database. so there's one for to
> questions and another for to options, that gives:
>
> {% for question in questionnaire %}
> <tr><td colspan="3"><b>{{forloop.counter}}. {{ question }}</b></
> td></tr>
> <tr>
> {% for alternative in question.get_alternatives %}
> <td><input type="radio" name="{{question.id}}"
> value="{{alternative.id}}"> {{ alternative }}</radio>
> {% endfor %}
> </tr>
> {% endfor %}
>
> The problem: depending on the number of questions, it can get pretty
> heavy and time-consuming to retrieve all questionnaire from the
> database.
>
> My raw solution was to copy the output html code of all questionnaires
> and save it on another "static" template,
> I'm aware of the non-practicality of such so, here's my question:
>
> Is there any pythonic-djangonic way to, in a lazy-loading style,
> generate a template file where all the database retrieving was already
> done and the whole html code was made, only regenerating this file if
> something was changed in the database?
>
> Did I made this question understandable? (I'm sorry, I'm a noob)
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