how to iterate through dictionary in a dictionary in django template?

155,764

Solution 1

Lets say your data is -

data = {'a': [ [1, 2] ], 'b': [ [3, 4] ],'c':[ [5,6]] }

You can use the data.items() method to get the dictionary elements. Note, in django templates we do NOT put (). Also some users mentioned values[0] does not work, if that is the case then try values.items.

<table>
    <tr>
        <td>a</td>
        <td>b</td>
        <td>c</td>
    </tr>

    {% for key, values in data.items %}
    <tr>
        <td>{{key}}</td>
        {% for v in values[0] %}
        <td>{{v}}</td>
        {% endfor %}
    </tr>
    {% endfor %}
</table>

Am pretty sure you can extend this logic to your specific dict.


To iterate over dict keys in a sorted order - First we sort in python then iterate & render in django template.

return render_to_response('some_page.html', {'data': sorted(data.items())})

In template file:

{% for key, value in data %}
    <tr>
        <td> Key: {{ key }} </td> 
        <td> Value: {{ value }} </td>
    </tr>
{% endfor %}

Solution 2

This answer didn't work for me, but I found the answer myself. No one, however, has posted my question. I'm too lazy to ask it and then answer it, so will just put it here.

This is for the following query:

data = Leaderboard.objects.filter(id=custom_user.id).values(
    'value1',
    'value2',
    'value3')

In template:

{% for dictionary in data %}
  {% for key, value in dictionary.items %}
    <p>{{ key }} : {{ value }}</p>
  {% endfor %}
{% endfor %}

Solution 3

If you pass a variable data (dictionary type) as context to a template, then you code should be:

{% for key, value in data.items %}
    <p>{{ key }} : {{ value }}</p> 
{% endfor %}
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darren
Author by

darren

Updated on July 08, 2022

Comments

  • darren
    darren almost 2 years

    My dictionary looks like this(Dictionary within a dictionary):

    {'0': {
        'chosen_unit': <Unit: Kg>,
        'cost': Decimal('10.0000'),
        'unit__name_abbrev': u'G',
        'supplier__supplier': u"Steve's Meat Locker",
        'price': Decimal('5.00'),
        'supplier__address': u'No\r\naddress here',
        'chosen_unit_amount': u'2',
        'city__name': u'Joburg, Central',
        'supplier__phone_number': u'02299944444',
        'supplier__website': None,
        'supplier__price_list': u'',
        'supplier__email': u'[email protected]',
        'unit__name': u'Gram',
        'name': u'Rump Bone',
    }}
    

    Now I'm just trying to display the information on my template but I'm struggling. My code for the template looks like:

    {% if landing_dict.ingredients %}
      <hr>
      {% for ingredient in landing_dict.ingredients %}
        {{ ingredient }}
      {% endfor %}
      <a href="/">Print {{ landing_dict.recipe_name }}</a>
    {% else %}
      Please search for an ingredient below
    {% endif %}
    

    It just shows me '0' on my template?

    I also tried:

    {% for ingredient in landing_dict.ingredients %}
      {{ ingredient.cost }}
    {% endfor %}
    

    This doesn't even display a result.

    I thought perhaps I need to iterate one level deeper so tried this:

    {% if landing_dict.ingredients %}
      <hr>
      {% for ingredient in landing_dict.ingredients %}
        {% for field in ingredient %}
          {{ field }}
        {% endfor %}
      {% endfor %}
      <a href="/">Print {{ landing_dict.recipe_name }}</a>
    {% else %}
      Please search for an ingredient below
    {% endif %}
    

    But this doesn't display anything.

    What am I doing wrong?

  • darren
    darren over 12 years
    thanks for your answer. I have recipe_name one level up and didn't show that level of the dictionary. Thank you for your answer! I couldn't use values[0] instead I had to values.items
  • Srikar Appalaraju
    Srikar Appalaraju over 12 years
    cool! glad to share what i knew. Code was typed free hand, so some mistakes are inevitable.
  • Dave
    Dave almost 9 years
    Thanks for mentioning .items. The documentation https://docs.djangoproject.com/en/1.4/topics/templates/ gives an example that doesn't work, but no example that does work. {% for k,v in dict %} gives bizarre results - k is the first character of every key and v is blank, while {% for k in dict %} returns the full key but with no way to retrieve values (since dict.k treats k as a literal character).
  • Dave
    Dave almost 9 years
    Upvoted :-). For future readers, the tags and filters reference https://docs.djangoproject.com/en/1.8/ref/templates/builtins‌​/#for documents this, along with the forloop variables that can be helpful when formatting a list.
  • Srikar Appalaraju
    Srikar Appalaraju almost 8 years
    @Flimm dictionary is unsorted. if you want sorted order for key in sorted(mydict.iterkeys()): print "%s: %s" % (key, mydict[key])
  • Kevin_TA
    Kevin_TA almost 7 years
    values.items +1