Couldn't you do:
test = []
test.append({'title': 'My First Title', 'post': 'My First Post'});
test.append({'title': 'My Second Title', 'post': 'My Second Post'});
?
On Jun 18, 2008, at 4:11 PM, Juan Hernandez wrote:
> it does not return any values.
>
> I did post.0.title because the
Oops, I totally missed that 'test' was a dictionary. This should work
then:
{% for key, x in posts.items %}
{{ x.title }}
{{ x.post }}
{% endfor %}
Alternately, if your keys are just going to be sequential integers,
you could just put this stuff into a list and use my first suggestion.
On
Juan Hernandez wrote:
> it does not return any values.
>
> I did post.0.title because the dictionary is established like this:
>
> >>> test
> {0: {'post': 'My First Post', 'title': 'My First Title'}, 1: {'post':
> 'My Second Post', 'title': 'My Second Title'}}
>
> and it goes on and on
>
>
it does not return any values.
I did post.0.title because the dictionary is established like this:
>>> test
{0: {'post': 'My First Post', 'title': 'My First Title'}, 1: {'post': 'My
Second Post', 'title': 'My Second Title'}}
and it goes on and on
If I could iterate over it, like post.0.title,
I could be completely mistaken, but can't you replace 'post.0.title'
and 'post.0.post' with 'x.title' and 'x.post', respectively?
On Jun 17, 1:50 pm, "Juan Hernandez" <[EMAIL PROTECTED]> wrote:
> Hey there
>
> I've been practicing for a while with the template system and I have some
> questions:
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