django1/django/views/csrf.py

65 lines
2.2 KiB
Python

from django.http import HttpResponseForbidden
from django.template import Context, Template
from django.conf import settings
# We include the template inline since we need to be able to reliably display
# this error message, especially for the sake of developers, and there isn't any
# other way of making it available independent of what is in the settings file.
CSRF_FAILRE_TEMPLATE = """
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">
<html lang="en">
<head>
<meta http-equiv="content-type" content="text/html; charset=utf-8">
<title>403 Forbidden</title>
</head>
<body>
<h1>403 Forbidden</h1>
<p>CSRF verification failed. Request aborted.</p>
{% if DEBUG %}
<h2>Help</h2>
{% if reason %}
<p>Reason given for failure:</p>
<pre>
{{ reason }}
</pre>
{% endif %}
<p>In general, this can occur when there is a genuine Cross Site Request Forgery, or when
<a
href='http://docs.djangoproject.com/en/dev/ref/contrib/csrf/#ref-contrib-csrf'>Django's
CSRF mechanism</a> has not been used correctly. For POST forms, you need to
ensure:</p>
<ul>
<li>The view function uses <a
href='http://docs.djangoproject.com/en/dev/ref/templates/api/#subclassing-context-requestcontext'><code>RequestContext</code></a>
for the template, instead of <code>Context</code>.</li>
<li>In the template, there is a <code>{% templatetag openblock %} csrf_token
{% templatetag closeblock %}</code> template tag inside each POST form that
targets an internal URL.</li>
</ul>
<p>You're seeing the help section of this page because you have <code>DEBUG =
True</code> in your Django settings file. Change that to <code>False</code>,
and only the initial error message will be displayed. </p>
<p>You can customize this page using the CSRF_FAILURE_VIEW setting.</p>
{% else %}
<p><small>More information is available with DEBUG=True.</small></p>
{% endif %}
</body>
</html>
"""
def csrf_failure(request, reason=""):
"""
Default view used when request fails CSRF protection
"""
t = Template(CSRF_FAILRE_TEMPLATE)
c = Context({'DEBUG': settings.DEBUG,
'reason': reason})
return HttpResponseForbidden(t.render(c), mimetype='text/html')