Django template variables as key for a list

Question

i've this context created in my view:

{'cats': {'LIST1': ['a','b','c'], 'LIST2': ['aa','bb','cc','dd'], 'LIST3': ['f','g']}}

what i want to do is to print the list title and then all the items, something like

LIST1:
- a
- b
- c

for all the lists, so i did this in my template

{% for  l_name, l in cats %}
    {{ l_name }}

    {%for lv in l %}
        {{ lv }}
    {% endfor %}
{% endfor %}

and this prints only the list names, without printing out the list. where's the mistake?

thanks

mmm, i don't think i can use [] in templates. anyway i found the problem. i need to create touple with () not {} — EsseTi
– EsseTi, Commented Nov 17, 2012 at 13:35

lqc · Accepted Answer · 2012-11-17 13:36:05Z

5

If you want to iterate over keys and values, you can use:

{% for name, lst in cats.iteritems %}
.... 
{% endfor %}

This just calls the iteritems method of a dictionary, which returns an iterator over a list of 2-tuples.

answered Nov 17, 2012 at 13:36

lqc

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Comments

EsseTi · Accepted Answer · 2012-11-17 13:38:08Z

0

Just for the record, the problem is how the data are created. so instead of

{'cats': {'LIST1': ['a','b','c'], 'LIST2': ['aa','bb','cc','dd'], 'LIST3': ['f','g']}}

i need a list of touple so:

{'cats': [('LIST1', ['a','b','c']), ('LIST2', ['aa','bb','cc','dd']), ('LIST3', ['f','g'])]}

answered Nov 17, 2012 at 13:38

EsseTi

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