contextmenu.py 文件源码

python
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项目:linkchecker-gui 作者: linkcheck 项目源码 文件源码
def can_view_parent_source (self, url_data):
        """Determine if parent URL source can be retrieved."""
        if not url_data.valid:
            return False
        parent = url_data.parent_url
        if not parent:
            return False
        # Directory contents are dynamically generated, so it makes
        # no sense in viewing/editing them.
        if parent.startswith(u"file:"):
            path = urlparse.urlsplit(parent)[2]
            return not os.path.isdir(get_os_filename(path))
        if parent.startswith((u"ftp:", u"ftps:")):
            path = urlparse.urlsplit(parent)[2]
            return bool(path) and not path.endswith(u'/')
        # Only HTTP left
        return parent.startswith((u"http:", u"https:"))
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