Python 使用StaticGenerator从Django项目生成静态HTML

我正在尝试为我在Django中构建的站点生成静态HTML。到目前为止我想到的是。在我看来，它应该类似于github页面中的示例，但我已经编写了下面的代码，它给出了下面的错误

代码：

错误（为简洁起见进行了编辑）：

---

好的，在查看您的跟踪后，我将猜测您的

设置。WEB\u ROOT

是一个元组

相关线路：

# web root set
self.web_root = getattr(self.settings, 'WEB_ROOT')

# web root used 
filename = self.fs.join(self.web_root, path.lstrip('/')).encode('utf-8')

# breaks on real os.path.join
return os.path.join(paths[0], *[path.lstrip("/") for path in paths[1:]])

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哇，那是个愚蠢的错误。我让WEB_根在末尾用逗号书写。我总是被标点符号绊倒。：-）非常感谢。我希望我能投你一票，但是声誉系统把我拒之门外。太棒了！很高兴你发现了问题。附言：嘘。甚至不能接受？我想知道是不是因为我在某一点上删除了这篇文章。谢谢：）那么它在upvote上说了什么？在这一点上我只是好奇，因为我从来没有被阻止投票。投票需要15%的声誉

Traceback (most recent call last):
    quick_publish('/')
  File "/usr/local/lib/python2.6/dist-packages/staticgenerator/__init__.py", line 232, in quick_publish
    return StaticGenerator(*resources).publish()
  File "/dist-packages/staticgenerator/__init__.py", line 229, in publish
    return self.do_all(self.publish_from_path)
  File "/dist-packages/staticgenerator/__init__.py", line 223, in do_all
    return [func(path) for path in self.resources]
  File "/dist-packages/staticgenerator/__init__.py", line 187, in publish_from_path
    filename, directory = self.get_filename_from_path(path)
  File "/dist-packages/staticgenerator/__init__.py", line 179, in get_filename_from_path
    filename = self.fs.join(self.web_root, path.lstrip('/')).encode('utf-8')
  File "/dist-packages/staticgenerator/filesystem.py", line 38, in join
    return os.path.join(paths[0], *[path.lstrip("/") for path in paths[1:]])
  File "/usr/lib/python2.6/posixpath.py", line 67, in join
    elif path == '' or path.endswith('/'):
AttributeError: 'tuple' object has no attribute 'endswith'

# web root set
self.web_root = getattr(self.settings, 'WEB_ROOT')

# web root used 
filename = self.fs.join(self.web_root, path.lstrip('/')).encode('utf-8')

# breaks on real os.path.join
return os.path.join(paths[0], *[path.lstrip("/") for path in paths[1:]])