Is there a way to remove an item from a list in the Django template language?
I have a situation where I'm iterating through one list, and printing the first item in another list. Once the first item is printed I want to remove it from that list.
See below:
{% for item in list1 %}
{{list2.0}}
#remove list2.0 from list2
{% endfor %}
If your list1 and list2 are indeed lists and not querysets, this seems to work:
{{ list2 }} {# show list2 #}
{% for item in list1 %}
{{ list2.0 }}
{# remove list2.0 from list2 #}
{{ list2.pop.0 }}
{% endfor %}
{{ list2 }} {# empty #}
Note that pop does not return in this case, so you still need {{ list2.0 }} explicitly.
I would try to filter out the item in the view if at all possible. Otherwise you can add in an if or if not statement inside the for loop.
{% for item in list%}
{% if item.name != "filterme" %}
{{ item.name }}
{% endif %}
{% endfor %}
You can't delete an item but you can get the list without a certain item (at a constant index)
{% with list2|slice:"1:" as list2 %}
...
{% endwith %}
Of course, nesting rules apply, etc.
In general, I you find yourself doing complex data structure manipulation, just move it to Python - it'd be faster and cleaner.
There is no such built-in template tag. I understand that you don't want to print first item of list2 if list1 is not empty. Try:
{% for item in list1 %}
{{list2.0}}
...
{% endfor %}
{% for item in list2 %}
{% if list1 and forloop.counter == 1 %}
# probably pass
{% else %}
{{ item }}
{% endif %}
{% endfor %}
This is not a good idea to manipulate the content of the list in templates.
ifstatment insideforloop body.popinside the template, provided your list is really a list, and not a queryset (otherwise, in your view, doqueryset = list(queryset). Then, in your template, try{{ list2.pop }}. I'm also not sure if you can provide an argument to pop, something along the lines of{{ list2.pop|forloop.counter }}. Finally, you could write your ownpoptag that does exactly this. I can probably come up with some code for that if you like.