Again, i'm a noob...
Thank you very much for pointing me in the right direction, I'll try
your recomendations :)
On 28 dez, 03:41, Mayuresh Phadke <mayur...@gmail.com> wrote:
> 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/geeroohttp://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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