if I have a variable in the context of unknown length, for example;list=[{'key':'A'},{'key':'B'},{'key':'C'}]
How can I get the last object? {{ list.0.key }}
works for the first, but {{ list.-1.key }}
gives;Could not parse the remainder: '-1.key' from 'list.-1.key'
What does {{ name }} this mean in Django Templates? Django. It will be displayed as name in HTML. The name will be replaced with values of Python variable. {{ name }} will be the output.
autoescape. Controls the current auto-escaping behavior. This tag takes either on or off as an argument and that determines whether auto-escaping is in effect inside the block. The block is closed with an endautoescape ending tag.
From the documentation: {% extends variable %} uses the value of variable. If the variable evaluates to a string, Django will use that string as the name of the parent template. If the variable evaluates to a Template object, Django will use that object as the parent template.
Django for loop counter All the variables related to the counter are listed below. forloop. counter: By using this, the iteration of the loop starts from index 1. forloop. counter0: By using this, the iteration of the loop starts from index 0.
Thanks everyone for you help, it lead me to the realisation that I can use the with tag.
{% with list|last as last %} {{ last.key }} {% endwith %}
Use the last
template tag:
{{ value|last }}
If value is the list
['a', 'b', 'c', 'd']
, the output will be the string"d"
.
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