title: TypeError: context must be a dict rather than Context
body: <p>I'm trying to build a search engine into a django blog application and when I ran the command:</p>
  
  <pre><code>&gt;&gt;&gt; manage.py build_solr_schema
  </code></pre>
  
  <p>I got this error:</p>
  
  <pre><code>Traceback (most recent call last):
    File "C:\Users\KOLAPO\Google Drive\Python\Websites\mysite\manage.py", line 22, in &lt;module&gt;
      execute_from_command_line(sys.argv)
    File "C:\Users\KOLAPO\Anaconda3\lib\site-packages\django\core\management\__init__.py", line 363, in execute_from_command_line
      utility.execute()
    File "C:\Users\KOLAPO\Anaconda3\lib\site-packages\django\core\management\__init__.py", line 355, in execute
      self.fetch_command(subcommand).run_from_argv(self.argv)
    File "C:\Users\KOLAPO\Anaconda3\lib\site-packages\django\core\management\base.py", line 283, in run_from_argv
      self.execute(*args, **cmd_options)
    File "C:\Users\KOLAPO\Anaconda3\lib\site-packages\django\core\management\base.py", line 330, in execute
      output = self.handle(*args, **options)
    File "C:\Users\KOLAPO\Anaconda3\lib\site-packages\haystack\management\commands\build_solr_schema.py", line 29, in handle
      schema_xml = self.build_template(using=using)
    File "C:\Users\KOLAPO\Anaconda3\lib\site-packages\haystack\management\commands\build_solr_schema.py", line 57, in build_template
      return t.render(c)
    File "C:\Users\KOLAPO\Anaconda3\lib\site-packages\django\template\backends\django.py", line 64, in render
      context = make_context(context, request, autoescape=self.backend.engine.autoescape)
    File "C:\Users\KOLAPO\Anaconda3\lib\site-packages\django\template\context.py", line 287, in make_context
      raise TypeError('context must be a dict rather than %s.' % context.__class__.__name__)
  TypeError: context must be a dict rather than Context.
  </code></pre>
  
  <p>What's wrong?</p>
  
  <p>Note: I'm using Solr and Django-haystack for the search engine</p>