TypeError: context must be a dict rather than Context
我正在尝试将搜索引擎构建到django博客应用程序中，当我运行命令时：
1 | >>> manage.py build_solr_schema |
我收到了这个错误：
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 | Traceback (most recent call last): File"C:\Users\KOLAPO\Google Drive\Python\Websites\mysite\manage.py", line 22, in <module> 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. |
怎么了？
注意：我正在使用Solr和Django-haystack作为搜索引擎
我认为这个问题已经被pull请求1504修复了，但是从那时起它看起来还没有发布。