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Python Folder.get_relative_path方法代码示例

本文整理汇总了Python中fswrap.Folder.get_relative_path方法的典型用法代码示例。如果您正苦于以下问题:Python Folder.get_relative_path方法的具体用法?Python Folder.get_relative_path怎么用?Python Folder.get_relative_path使用的例子?那么恭喜您, 这里精选的方法代码示例或许可以为您提供帮助。您也可以进一步了解该方法所在fswrap.Folder的用法示例。


在下文中一共展示了Folder.get_relative_path方法的1个代码示例,这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞,您的评价将有助于系统推荐出更棒的Python代码示例。

示例1: Node

# 需要导入模块: from fswrap import Folder [as 别名]
# 或者: from fswrap.Folder import get_relative_path [as 别名]

#.........这里部分代码省略.........
    """
    Represents any folder that is processed by hyde
    """

    def __init__(self, source_folder, parent=None):
        super(Node, self).__init__(source_folder)
        if not source_folder:
            raise HydeException("Source folder is required"
                                " to instantiate a node.")
        self.root = self
        self.module = None
        self.site = None
        self.source_folder = Folder(str(source_folder))
        self.parent = parent
        if parent:
            self.root = self.parent.root
            self.module = self.parent.module if self.parent.module else self
            self.site = parent.site
        self.child_nodes = []
        self.resources = []

    def contains_resource(self, resource_name):
        """
        Returns True if the given resource name exists as a file
        in this node's source folder.
        """

        return File(self.source_folder.child(resource_name)).exists

    def get_resource(self, resource_name):
        """
        Gets the resource if the given resource name exists as a file
        in this node's source folder.
        """

        if self.contains_resource(resource_name):
            return self.root.resource_from_path(
                self.source_folder.child(resource_name))
        return None

    def add_child_node(self, folder):
        """
        Creates a new child node and adds it to the list of child nodes.
        """

        if folder.parent != self.source_folder:
            raise HydeException("The given folder [%s] is not a"
                                " direct descendant of [%s]" %
                                (folder, self.source_folder))
        node = Node(folder, self)
        self.child_nodes.append(node)
        return node

    def add_child_resource(self, afile):
        """
        Creates a new resource and adds it to the list of child resources.
        """

        if afile.parent != self.source_folder:
            raise HydeException("The given file [%s] is not"
                                " a direct descendant of [%s]" %
                                (afile, self.source_folder))
        resource = Resource(afile, self)
        self.resources.append(resource)
        return resource

    def walk(self):
        """
        Walks the node, first yielding itself then
        yielding the child nodes depth-first.
        """
        yield self
        for child in sorted([node for node in self.child_nodes]):
            for node in child.walk():
                yield node

    def rwalk(self):
        """
        Walk the node upward, first yielding itself then
        yielding its parents.
        """
        x = self
        while x:
            yield x
            x = x.parent

    def walk_resources(self):
        """
        Walks the resources in this hierarchy.
        """
        for node in self.walk():
            for resource in sorted([resource for resource in node.resources]):
                yield resource

    @property
    def relative_path(self):
        """
        Gets the path relative to the root folder (Content, Media, Layout)
        """
        return self.source_folder.get_relative_path(self.root.source_folder)
开发者ID:LestherSK,项目名称:hyde,代码行数:104,代码来源:site.py


注:本文中的fswrap.Folder.get_relative_path方法示例由纯净天空整理自Github/MSDocs等开源代码及文档管理平台,相关代码片段筛选自各路编程大神贡献的开源项目,源码版权归原作者所有,传播和使用请参考对应项目的License;未经允许,请勿转载。