python xml-為什么find（“ ..”）不返回root

Question

為什么當root.findall("*")返回element時element.find("..")返回root ？

def XML_Extract_Node_Tags(Tree, Node_Tags):
    """
    :param Tree: xml.etree.ElementTree
    :param Node_Tags: list
    :return: ReturnVal:
    """

    for el in Tree.findall("//"):
        if el.tag not in Node_Tags:
            print(el.tag)
            # Need to remove the element and set its children equal to parent
            for subel in el.findall("*"):
                ## Add subel to grandparent (if exists)
                grand_parent = subel.find('../..')
                if grand_parent:
                    # If it has a grand parent
                    grand_parent.append(subel)
            # Remove el from tree
            if not el.find(".."):
                print(el.tag, el.attrib)
            else:
                el.find("..").remove(el)

    ReturnVal = Tree
    return ReturnVal

XML文件的前5行

<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE build SYSTEM "build.dtd">
<build id="58" localAgents="true" cm="usl000348:80" start="6/1/2016 3:31:19 PM">
<properties>
<property name="CommandLine">emake all --emake-annodetail=waiting,registry,md5,lookup,history,file,env --emake-annofile=../Emake-2agents-1st.xml --emake-root=../</property>

Answer 1

Python的xml.etree.ElementTree實現不記錄Element的父級。 因此， XPath的文檔包括以下內容：

..選擇父元素。 如果路徑嘗試到達start元素的祖先（調用元素find），則返回None 。