Django template on Google App Engine with if statement

Question

I'm pretty new to google app engine, and have just finished reading through the getting started. So I started implementing a simple app on GAE and soon got an error saying "TemplateSyntaxError, "'if' statement improperly formatted"" when I simply rendered an html which uses if statement. I solved the problem it using ifequal statement instead, but this tutorial shows me using if statement in the index.html (http://code.google.com/appengine/docs/python/gettingstarted/templates.html)

Do I miss anything to use if statement in a template?

Thanks, Yoo

UPDATED:
Here is details of the error. I think Chris' answer will let me use "Smart" if tag though. I'll try to update Django version to 1.2 soon.

Traceback (most recent call last):
  File "C:\Program Files\Google\google_appengine\google\appengine\ext\webapp\__init__.py", line 634, in __call__
    handler.get(*groups)
  File "C:\Store house\gae\community\src\community.py", line 24, in get
    self.response.out.write(template.render(path, template_values))
  File "C:\Program Files\Google\google_appengine\google\appengine\ext\webapp\template.py", line 72, in render
    t = load(template_path, debug)
  File "C:\Program Files\Google\google_appengine\google\appengine\ext\webapp\template.py", line 100, in load
    template = django.template.loader.get_template(file_name)
  File "C:\Program Files\Google\google_appengine\lib\django_0_96\django\template\loader.py", line 80, in get_template
    template = get_template_from_string(source, origin, template_name)
  File "C:\Program Files\Google\google_appengine\lib\django_0_96\django\template\loader.py", line 88, in get_template_from_string
    return Template(source, origin, name)
  File "C:\Program Files\Google\google_appengine\lib\django_0_96\django\template\__init__.py", line 158, in __init__
    self.nodelist = compile_string(template_string, origin)
  File "C:\Program Files\Google\google_appengine\lib\django_0_96\django\template\__init__.py", line 174, in compile_string
    return parser.parse()
  File "C:\Program Files\Google\google_appengine\lib\django_0_96\django\template\__init__.py", line 273, in parse
    compiled_result = compile_func(self, token)
  File "C:\Program Files\Google\google_appengine\lib\django_0_96\django\template\loader_tags.py", line 154, in do_extends
    nodelist = parser.parse()
  File "C:\Program Files\Google\google_appengine\lib\django_0_96\django\template\__init__.py", line 273, in parse
    compiled_result = compile_func(self, token)
  File "C:\Program Files\Google\google_appengine\lib\django_0_96\django\template\loader_tags.py", line 132, in do_block
    nodelist = parser.parse(('endblock', 'endblock %s' % block_name))
  File "C:\Program Files\Google\google_appengine\lib\django_0_96\django\template\__init__.py", line 273, in parse
    compiled_result = compile_func(self, token)
  File "C:\Program Files\Google\google_appengine\lib\django_0_96\django\template\defaulttags.py", line 655, in do_if
    raise TemplateSyntaxError, "'if' statement improperly formatted"
TemplateSyntaxError: 'if' statement improperly formatted

UPDATED 2:
According to this article(http://code.google.com/appengine/docs/python/tools/libraries.html#Django), current google app engine is already included Django 1.2 but current default version is 0.96, that's why I couldn't use if tag. To use the version 1.2, follow an instruction in the link above. Now, I can use "Smart" if tag. Thanks everyone :)

could you show the excerpt from your template that raises the error? — arie

Chris Chris · Accepted Answer · 2011-04-15T13:57:20

If you tried to use the tag:

{% if x == 1 %}

and you are getting an error that is resolved with

{% ifequal x 1 %}

that is a sign your version of Django is 1.1 or lower. "Smart" if tags came with version 1.2

Edit to add, Django 1.2 and higher does work on GAE. See this blog post for how to set it up if you are indeed running an old version of Django.

Django template on Google App Engine with if statement

2 Answers