What if this filter is being used inside a loop and node is changing all
the time, which happens to be the case?
Another loop just to find out if a node is inside a list of nodes is not
only slower and verbose but also a dumb method, since we can't even
break as soon as we have a positive response.
Compare:
{% for _node in open_nodes %}
{% ifequal node.id _node.id %}
{{ node.name }}
{% endif %}
{% endfor %}
with:
{% if node.id|in:open_nodes %}
{{ node.name }}
{% endif %}
I'm aware that this could have been done at the logic level by iterating
through all the nodes and adding a new bool key stating if they were
opened, but IMHO using the in keyword inside the templates doesn't break
the layer separation that django is trying to enforce.
My 2 cents :)
João
Russell Keith-Magee wrote:
> On 10/23/06, João Cruz Morais <[EMAIL PROTECTED]> wrote:
>
>>Hi there,
>>
>>Some time ago I've submitted this filter -
>>http://code.djangoproject.com/ticket/2741 - because i thought it was too
>>simple and obvious to be left outside the builtin pack.
>>Adrian rejected the purposal, and never said why, so I'm wondering, do
>>you know any other way to achieve this filter objective without, of
>>course, resorting to new filters/tags/etc?
>
>
> I haven't spoken to Adrian about his reasons, but I suspect I know why
> he rejected this proposal, and I would have rejected it for the same
> reason.
>
> Django is very explicit about templates not containing view logic. If
> you are checking set/list membership in a template, it indicates that
> there is some aspect of the view that you are not adequately capturing
> in your context.
>
> The right way to do this would be to put the 'IN' condition as a
> context variable (e.g., is_open: node.id in open_nodes), and then use
> a standard 'if' template tag (e.g., {%if is_open %}).
>
> This maintains the separation of template from view; you can modify
> the conditions under which the is_open block is displayed without
> having to modify the template.
>
> Yours,
> Russ Magee %-)
>
> >
>
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