Obtain the first part of an URL from Django template

Question

I use request.path to obtain the current URL. For example if the current URL is "/test/foo/baz" I want to know if it starts with a string sequence, let's say /test. If I try to use:

{% if request.path.startswith('/test') %}
    Test
{% endif %} 

I get an error saying that it could not parse the remainder of the expression:

Could not parse the remainder: '('/test')' from 'request.path.startswith('/test')'
Request Method: GET
Request URL:    http://localhost:8021/test/foo/baz/
Exception Type: TemplateSyntaxError
Exception Value:    
Could not parse the remainder: '('/test')' from 'request.path.startswith('/test')'
Exception Location: C:\Python25\lib\site-packages\django\template\__init__.py in   __init__, line 528
Python Executable:  C:\Python25\python.exe
Python Version: 2.5.4
Template error

One solution would be to create a custom tag to do the job. Is there something else existing to solve my problem? The Django version used is 1.0.4.

Answer 1

You can use the slice filter to get the first part of the url

{% if request.path|slice:":5" == '/test' %}
    Test
{% endif %} 

Cannot try this now, and don't know if filters work inside 'if' tag, if doesn't work you can use the 'with' tag

{% with request.path|slice:":5" as path %}
  {% if path == '/test' %}
    Test
  {% endif %} 
{% endwith %} 

Answer 2

Instead of checking for the prefix with startswith, you can get the same thing by checking for membership with the builtin in tag.

{% if '/test' in request.path %}
    Test
{% endif %} 

This will pass cases where the string is not strictly in the beginning, but you can simply avoid those types of URLs.