Django template get element in list of dictionary by index

Question

I have the following:

item0 = [{'itemCode': 'AZ001', 'price': 15.52}, {'itemCode': 'AB01', 'price': 31.2}, {'itemCode': 'AP01', 'price': 1.2}]
item1 = [{'itemCode': 'BZ001', 'price': 12.55}, {'itemCode': 'BB01', 'price': 34.1}]

In django template I would like to display the price of the elements of each list by index: 15.52, 12.55 then 31.2, 34.1 then 1.2

List sizes might not be equal, so I am sending the size of the largest list.

Iterating over the max list size:

{{i.item|index:forloop.counter0}} gets me {'itemCode': 'AZ001', 'price': 15.52}

If I want price, what can I do?

Doing {{i.item|index:forloop.counter0.price}} is giving me invalid key price at index 0.

In other words, I am sending elements in column order and would like to display them in row order without doing list comprehension using zip on the server.

Any solution?

You probably need to give more context. Normally instead of using a template tag to get an index based on forloop.counter0, you would iterate directly through item: `{% for elem in i.item %}... {{ elem.price }}`. Why can't you do that? — Daniel Roseman, Feb 26 '18 at 15:58
I have multiple lists that I am comparing next to each other. so I need to display them in row based, when I am sending them column based. I can do zip on the server, but I am trying other ways before going to zip. — Syntax_Error, Feb 26 '18 at 15:59
Also if you trying to access first element of list then you can do: `item.0.price`. — Shashank, Feb 26 '18 at 16:00
You could write a [custom template tag](https://stackoverflow.com/a/29664945/113962). However using `zip` (or `zip_longest` for lists with different lengths) seems like a better approach to me. — Alasdair, Feb 26 '18 at 16:18

Yannic Hamann · Answer 1 · 2018-02-26T21:20:45.380

Not sure if I get your question right, but this is the code what you are asking for.

views.py:

def get_context_data(self, **kwargs):
        context = super().get_context_data(**kwargs)
        context['item'] = [{'itemCode': 'AZ001', 'price': 15.52}, {'itemCode': 'AB01', 'price': 31.2}]
        return context

template.html:

{{ item.0.price }}

Results in 15.52

If you want to loop over it you can do it like so:

{% for i in item %}
    {{ i.price }}
{% endfor %}

After you have updated the question, I would do the following:

def get_context_data(self, **kwargs):
    context = super().get_context_data(**kwargs)
    item0 = [{'itemCode': 'AZ001', 'price': 15.52}, {'itemCode': 'AB01', 'price': 31.2}, {'itemCode': 'AP01', 'price': 1.2}]
    item1 = [{'itemCode': 'BZ001', 'price': 12.55}, {'itemCode': 'BB01', 'price': 34.1}]
    import itertools
    context['zip_longest'] = itertools.zip_longest(item0, item1)
    return context

template.html:

{% for element in zip_longest %}
    {% for item in element %}
        {% if item %}
            {{ item.price }} <br>
        {% endif %}
    {% endfor %}
{% endfor %}

Results in:

In my opinion, it's nothing wrong with using zip_longest, since it yields the values from a generator.

Edited the questions to explain more – Syntax_Error Feb 26 '18 at 16:06 — Syntax_Error, Feb 26 '18 at 16:06

score 0 · Answer 2 · edited Mar 18 '19 at 08:29

0

<ul>
     {% for key, value in dictionary.items %}
     <li><a href="{{key}}">{{value}}</a></li>
     {% endfor %}
</ul>

try using this, reference

edited Mar 18 '19 at 08:29

wscourge

10,657
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answered Mar 18 '19 at 08:08

mh-samiei

111
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Django template get element in list of dictionary by index

2 Answers2