AbstractProblemFactory
AbstractProblemFactory

Reputation: 9811

How to format dateTime in django template?

That:

{{ wpis.entry.lastChangeDate|date:"D d M Y" }}

gives me (why?):

2009-07-24 21:45:38.986156 

And, I don't know how to skip fraction part...

In my model, I have:

addedDate = models.DateTimeField(default=datetime.now)

Upvotes: 83

Views: 152215

Answers (5)

In addition to other answers, time filter below can only show a formatted time:

{{ wpis.entry.lastChangeDate|time:'H:i' }}

Upvotes: 3

Rodrigo Camargos
Rodrigo Camargos

Reputation: 142

{{ wpis.entry.lastChangeDate|date:"SHORT_DATETIME_FORMAT" }}
Replace SHORT_DATETIME_FORMAT with the date and/or time format you desire to use.

Here's a list of available date and time formats to use in Django templates : https://docs.djangoproject.com/en/4.0/ref/templates/builtins/#date https://docs.djangoproject.com/en/4.0/ref/templates/builtins/#time

Make sure the filter is applied to the correct field.

Upvotes: 4

Rockystech
Rockystech

Reputation: 3086

This is exactly what you want. Try this:

{{ wpis.entry.lastChangeDate|date:'Y-m-d H:i' }}

Upvotes: 235

Coc B.
Coc B.

Reputation: 665

You can use this:

addedDate = datetime.now().replace(microsecond=0)

Upvotes: 10

rob
rob

Reputation: 37684

I suspect wpis.entry.lastChangeDate has been somehow transformed into a string in the view, before arriving to the template.

In order to verify this hypothesis, you may just check in the view if it has some property/method that only strings have - like for instance wpis.entry.lastChangeDate.upper, and then see if the template crashes.

You could also create your own custom filter, and use it for debugging purposes, letting it inspect the object, and writing the results of the inspection on the page, or simply on the console. It would be able to inspect the object, and check if it is really a DateTimeField.

On an unrelated notice, why don't you use models.DateTimeField(auto_now_add=True) to set the datetime on creation?

Upvotes: 5

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