CODEWITHSUNDEEP
pythondjangoarraystemplatesindexing
janb
janb

Reputation: 136

How can I use a variable as index in django template?

How can I use a variable as index in django template?

Right now I get this error:

Exception Type:
TemplateSyntaxError

Exception Value:    
Could not parse the remainder: '[year]' from 'bikesProfit.[year]'

I also tried {{ bikesProfit.year }} but this gives an empty result.

 {%  for year in years_list %}

        <tr>
            <th>Total profit {{ year }}:</th>
        </tr>
        <tr>
            <th></th>
            <th> {{ bikesProfit[year] }} </th>

...

Upvotes: 7

Views: 11917

Answers (3)

Serhii Holinei
Serhii Holinei

Reputation: 5864

It's a very common question, there are a lot of answers on SO.

You can make custom template filter:

@register.filter
def return_item(l, i):
    try:
        return l[i]
    except:
        return None

Usage:

{{ bikesProfit|return_item:year }}

Upvotes: 21

Martin Maillard
Martin Maillard

Reputation: 2811

I don't think this is directly possible (edit: you can manually implement it with filters, as shown in goliney's answer). The documentation says:

Because Django intentionally limits the amount of logic processing available in the template language, it is not possible to pass arguments to method calls accessed from within templates. Data should be calculated in views, then passed to templates for display.

If your case is not more complicated than what you're showing, the best solution in my opinion would be to loop over bikeProfit instead.

{% for year, profit in bikeProfit.items %}
    ...
    <th>Total profit {{ year }}:</th>
    ...
    <th> {{ profit }} </th>
    ...

Upvotes: 3

Jon Clements
Jon Clements

Reputation: 142156

It's possible, but goes against the grain of the Django model...

If you have items of interest, then ideally your view should return an object just containing those items, (OTTOMH something like):

Something.objects.filter(something__typename='Bike', year__range=(1998, 2001))

Upvotes: 2

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