I have two list objects of the same length with complementary data i want to render is there a way to render both at the same time ie.
{% for i,j in table, total %}
{{ i }}
{{ j }}
{% endfor %}
or something similar?
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Does this answer your question? Iterating through two lists in Django templatesdjvg– djvg2023-05-08 09:59:39 +00:00Commented May 8, 2023 at 9:59
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6 Answers
If both lists are of the same length, you can return zipped_data = zip(table, total) as template context in your view, which produces a list of 2-valued tuples.
Example:
>>> lst1 = ['a', 'b', 'c']
>>> lst2 = [1, 2, 3]
>>> zip(lst1, lst2)
[('a', 1), ('b', 2), ('c', 3)]
In your template, you can then write:
{% for i, j in zipped_data %}
{{ i }}, {{ j }}
{% endfor %}
Also, take a look at Django's documentation about the for template tag here. It mentions all possibilities that you have for using it including nice examples.
2 Comments
mike
this question seemed rather easy, or im rather dum. but thanks to every one that answered!
pemistahl
@mike You are not dumb. All of us were at this point some time ago when learning about Python and Django. And Django's docs don't mention Python's
zip function, so this question was definitely useful. I'm glad that I could help you. :)For any recent visitors to this question, forloop.parentloop can mimic the zipping of two lists together:
{% for a in list_a %}{% for b in list_b %}
{% if forloop.counter == forloop.parentloop.counter %}
{{a}} {{b}}
{% endif %}
{% endfor %}{% endfor %}
Comments
Use python's zip function and zip the 2 lists together.
In your view:
zip(table, list)
In your template, you can iterate this like a simple list, and use the .0 and .1 properties to access the data from table and list, respectively.
Comments
If it's just the variables i and j that you're looking at then this should work -
return render_to_response('results.html',
{'data': zip(table, list)})
{% for i, j in data %}
<tr>
<td> {{ i }}: </td> <td> {{ j }} </td>
</tr>
{% endfor %}
(credit to everyone else who answered this question)
2 Comments
Jordan
Converting from a list to a dict and calling sorted on it will likely change the sorting order of the data.
Aidan Ewen
good point, I've updated my answer to simply iterate the zip.
Rather than using a dictionary (which does not guarantee any kind of sorting), use the python zip function on the two lists and pass it to the template.
Comments
You'll have to do this in the view - use the builtin zip function to make a list of tuples, then iterate over it in the template.
Template logic is purposely simple, anything even slightly complex should be done in the view.
