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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