代码之家 › 专栏 › 技术社区 › fredrik

具有页属性sortIndex的树遍历

tree-traversal google-app-engine python

fredrik · 技术社区 · 16 年前

我正在根据从 db.Model

每个页面条目都有一个parentKey属性,它是 db.SelfReferenceProperty() 和一个 db.IntegerProperty() 叫做sortIndex。

pages = Pages.gql('ORDER BY sortIndex').fetch(1000)
build_tree(pages)

以及构建树:

def build_tree(nodes, *args):
    # create empty tree to fill
    tree = {}
    build_tree_recursive(tree, None, nodes, *args)

    return tree

def build_tree_recursive(tree, parent, nodes, *args):
    # find root children, first level nodes have no parentKey
    if parent is None:
        children  = [n for n in nodes if n.parentKey == None]
    # find children
    else:
        children  = [n for n in nodes if n.parentKey is not None and n.parentKey.key() == parent]

    # build a subtree for each child
    for child in children:
        # start new subtree
        key = child.key()
        # Use page entry key as unique dict key
        tree[key] = { 'page' : child, 'children' : {}}
        # call recursively to build a subtree for current node
        build_tree_recursive(tree[key]['children'], key, nodes)

问题是列表get被重新排列并且不遵循det ORDER BY。我认为这是因为每一页都被放在列表中,当找到一个合适的父母。但即使是第一级(有 parentKey == None

我曾尝试在树[str(I)+'\'+str(key)]上使用循环计数器设置前缀,但仍然没有按正确的顺序返回。

所以问题是如何把它们排列整齐?

编辑[已解决]:

1 回复 | 直到 13 年前

fredrik 13 年前

为了保留作为参数发送的列表的顺序来构建树,我移到了另一个角度。我改用列表,顺序不变:

def build_tree(nodes, *args):
    # create empty tree to fill
    t = {}
    # First group all pages w/ same parent
    for node in nodes:
        if node.parentKey is None:
            key = 'root'
        else:
            key = node.parentKey.key()

        if not t.has_key(key):
            t[key] = []

        t[key].append({ 'page' : node, 'children' : []})

    pageTree = t['root']
    # Iterate over there
    build_page_tree(pageTree, t)

    return pageTree

def build_page_tree(pageTree, nodes):
    #Loop over selected list
    for parent, node in nodes.iteritems():
        # We don't need to loop over first level node
        if parent is not 'root':
            # Loop over current level in page tree
            for item in pageTree:
                # Match keys
                if item['page'].key() == parent:
                    # Save node as child
                    item['children'] = node
                    # Only need to loop over childs if they are present
                    build_page_tree(item['children'], nodes)