Think you're not undesrtanding...
Example:
python code:
myList = ["one", "two", "three"]
for item in myList: print item
output:
one
two
three
django template tag "for" can get the itens in list in same way... but
you get the itens and the instance variables of the model model:
class ImageModel:
name
description
links
{% for image in imageList %}
{{ image.name }}
{{ image.description }}
{{ image.links }}
{% endfor %}
In your case "i" cannot be an index of your list, but is the current
item of the list itself, a instance of your Model.
2011/11/25 marjenni <[email protected]>:
> Thank you for the quick reply :)
>
> What I am currently doing is passing 3 separate lists to the template,
> each representing the column of a table so that i can do something
> this:
>
> {% for i in numberOfImages %}
> <tr>
> <td> {{ images.i}} </td>
> <td> {{ descriptions.i}} </td>
> <td> {{ links.i}} </td>
>
> What is the best way to do this then?
>
> many thanks
>
> Mark
>
>
> On Nov 25, 1:25 pm, Tom Evans <[email protected]> wrote:
>> On Fri, Nov 25, 2011 at 1:18 PM, marjenni <[email protected]>
>> wrote:
>> > Hi all,
>> > Why does this not work?
>>
>> > {% for i in numberOfImages %}
>> > <tr>
>> > <td> {{ images.i}} </td>
>>
>> It doesn't work because there is no variable interpolation inside a
>> template tag. That snippet asks for the attribute named 'i' on the
>> 'images' variable.
>>
>> As an earlier poster said, pass a list of images to output, and then
>> iterate through that list. If you need the ordinal of where you are in
>> the list, that is made available by the for tag:
>>
>> https://docs.djangoproject.com/en/1.3/ref/templates/builtins/#for
>>
>> Cheers
>>
>> Tom
>
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