How to parse "2015-01-01T00:00:00Z" in Django Template?

Question

In my Django html template, I get my SOLR facet_date result using haystack in the format "2015-01-01T00:00:00Z". How can I parse it in format "01/01/2015" in my template? My template is

{{ facets.dates.created.start }}

What "|date:" option should I add to my template? Thanks!

Paulo Scardine · Accepted Answer · 2015-12-01T05:05:22.830

9

If your date is a ISO string instead of a Python datetime.datetime, I guess you will have to parse it on the view or write a custom filter:

# yourapp/templatetags/parse_iso.py
from django.template import Library
import datetime

register = Library()

@register.filter(expects_localtime=True)
def parse_iso(value):
    return datetime.datetime.strptime(value, "%Y-%m-%dT%H:%M:%SZ")

Then at the template:

{% load parse_iso %}

{{ value|parse_iso|date:'d/m/Y'}}

[edit]

got this error Exception Type: TemplateSyntaxError at /search/ Exception Value: 'parse_iso' is not a valid tag library: Template library parse_iso not found

Make sure you follow the code layout prescribed in the docs:

yourapp/
    __init__.py
    models.py
    ...
    templatetags/
        __init__.py
        parse_iso.py
    views.py

Your country may use m/d/Y (01/01/2015 is ambiguous, I suggest using an example like 31/01/2015 so it is clear if the first number represents day or month).

edited Dec 01 '15 at 05:05

answered Dec 01 '15 at 04:40

Paulo Scardine

73,447
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got this error `Exception Type: TemplateSyntaxError at /search/ Exception Value: 'parse_iso' is not a valid tag library: Template library parse_iso not found` – MC X Dec 01 '15 at 04:49
Please follow the directory [layout](https://docs.djangoproject.com/en/1.8/howto/custom-template-tags/#code-layout) described in Django docs: make sure `templatetags` is a python module inside your app containing a file named `__init__.py` (may be empty) along with `parse_iso.py`. – Paulo Scardine Dec 01 '15 at 04:50
Great answer. Though instead of `.strptime()` to manually formatting the string, it is better to use `.fromisoformat()`. – xyres Apr 20 '22 at 16:28

score 4 · Answer 2 · answered Dec 01 '15 at 04:42

If {{ facets.dates.created.start }} is a datetime object then you can use

{{ facets.dates.created.start|date:"SHORT_DATE_FORMAT" }}

In case you are providing a string you can create a template filter to convert the string to datetime object and apply the date filter

@register.filter
def stringformat(value, args):
    return datetime.strptime(value, args)

In the template:

{{ facets.dates.created.start|stringformat:"%Y-%m-%dT%H:%M:%SZ"|date:"SHORT_DATE_FORMAT" }}

shahram kalantari · Answer 3 · 2015-12-01T06:56:46.657

-1

You can use Django template tags for this. You need to use {{my_date|date:"some_format"}} which takes "my_date" as the argument (it should be a date object) to "date" tag and then formats it based on the given format.

{{facets.dates.created.start|date:"d/m/Y"}}

edited Dec 01 '15 at 06:56

answered Dec 01 '15 at 04:33

shahram kalantari

771
7
21

I tried this, but does not work. When I just do `{{ facets.dates.created.star }}`, 2015-01-01T00:00:00Z will be shown. But when I add `date:"d/m/Y"`, nothing shows when I load the template. – MC X Dec 01 '15 at 04:48
Please include some text explaining why you think this might solve the OP's problem. Help others to understand. – APC Dec 01 '15 at 05:27

How to parse "2015-01-01T00:00:00Z" in Django Template?

3 Answers3