error when django with rest-framework deployed

Question

I am having error when trying to open url of Django Rest framework. It was working fine locally, but when I deployed it on server, I am having following error. On server I have django 1.9.

Exception Value:    

'url' is not a valid tag or filter in tag library 'future'

Exception Location:     /home/maxo/django-trunk/django/template/base.py in parse, line 506

Error during template rendering

In template /usr/local/lib/python2.7/dist-packages/rest_framework/templates/rest_framework/base.html, error at line 1

'url' is not a valid tag or filter in tag library 'future'
1   

      {% load url from future %}



2   {% load staticfiles %}
3   {% load rest_framework %}
4   <!DOCTYPE html>
5   <html>
6       <head>
7           {% block head %}
8   
9               {% block meta %}
10                  <meta http-equiv="Content-Type" content="text/html; charset=utf-8"/>
11                  <meta name="robots" content="NONE,NOARCHIVE" />

NOTE : When I removed following line: {% load url from future %} from base.html Its working fine now, but then style of rest api is gone. Is there any other alternative to replace {% load url from future %}?

When I removed following line: {% load url from future %} from base.html Its working fine now, but then style of rest api is gone. Is there any other alternative to replace {% load url from future %}? — user3487775
– user3487775, Commented Jul 25, 2015 at 8:57

Rahul Gupta · Accepted Answer · 2015-07-25 09:02:28Z

4

In Django 1.9, url template tag was removed from the future template tag library.

From Django 1.9 release notes:

ssi and url template tags will be removed from the future template tag library (used during the 1.3/1.4 deprecation period).

So, now you can't load the url tag from the future library in Django 1.9. You can use the built-in url tag instead.

{% url 'some-url-name' %}

edited Jul 25, 2015 at 9:02

answered Jul 25, 2015 at 8:55

Rahul Gupta

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4 Comments

user3487775 Over a year ago

When I removed following line: {% load url from future %} from base.html Its working fine now, but then style of rest api is gone. Is there any other alternative to replace {% load url from future %}?

Rahul Gupta Over a year ago

You can use the default url tag instead. {% url 'some-url-name' %}

user3487775 Over a year ago

so do I need replace 'some-url-name' with anything? as I need to replace {% load url from future %} with something on base.html file which is in DJANGO rest Framwork folder

Rahul Gupta Over a year ago

some-url-name can be a view function or a reverse url name. Lets say if your url has a name=my_url_name argument defined in the urls.py file, then you can just use{% url 'my_url_name' %}. In case of namespaced URL, you need to specify the fully qualified URL like {% url 'myapp:view_name' %}

rrosa · Accepted Answer · 2016-05-17 04:41:10Z

1

you can fix that by installing a newer version of djangorestframework

pip install 'djangorestframework>=3.2.3'

i don't think it's a good idea to develop on a version of django and deploy on some other version, that'll most likely give you problems. i'd work using virtualenv, and saving a requirements.txt file with the version of all the packages you're using. that way when you deploy you can run:

pip install -r requirements.txt

and that'll install the same versions as you are using in your dev environment.

hope this helps. salú

answered May 17, 2016 at 4:41

rrosa

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