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按标签在每个元素后插入XML节点

xml python

JokerMartini · 技术社区 · 2 年前

我有以下xml,我想插入一个 <separatorItem/> 之后每 </subMenu> 这有一个 <scriptItem/> 之后。

或者换句话说,你可以重新措辞,因为我想插入一个 <分隔符项目/> 在xml层次结构的每个级别,在最后一个级别之后 <subMenu> 它在同一级别内包含更多元素。

<?xml version="1.0" encoding="UTF-8"?>
<mainMenu>
    <menuBar>
        <subMenu>
            <label>Main</label>
            <subMenu>
                <label>Sub Menu</label>
                <subMenu>
                    <label>Sub-Sub Menu</label>
                    <scriptItem>
                        <label>Email</label>
                    </scriptItem>
                </subMenu>
                <scriptItem>
                    <label>Manager</label>
                </scriptItem>
            </subMenu>
            <subMenu>
                <label>Dev</label>
                <scriptItem>
                    <label>Dialog</label>
                </scriptItem>
            </subMenu>
            <scriptItem>
                <label>Sample</label>
            </scriptItem>
            <scriptItem>
                <label>Home</label>
            </scriptItem>
            <scriptItem>
                <label>Browser</label>
            </scriptItem>
            <scriptItem>
                <label>Check</label>
            </scriptItem>
            <scriptItem>
                <label>Open</label>
            </scriptItem>
            <scriptItem>
                <label>Close</label>
            </scriptItem>
        </subMenu>
    </menuBar>
</mainMenu>

起始代码

import os
import re
import sys
import logging
import xml.dom.minidom
import xml.etree.ElementTree as ET


fp = os.path.abspath('elementTree.xml')
tree = ET.parse(fp)
root = tree.getroot()

# insert separator after every subMenu that's followed by a scriptItem
# parent = None
# for el in root.iter():
    # parent = root # find correct subMenu to place sep after
    # sep = ET.SubElement(parent, 'separatorItem')
sep = ET.Element('separatorItem')
root.insert(1, sep)


xmlstr = ET.tostring(root, encoding='UTF-8', method="xml")
xmlObj = xml.dom.minidom.parseString(xmlstr)
xmlstr = xmlObj.toprettyxml(encoding='UTF-8')

menuFilepath = os.path.abspath('updated.xml')
with open(menuFilepath, 'wb') as output:
    output.write(xmlstr)

最终目标看起来像这样。。。

<?xml version="1.0" encoding="UTF-8"?>
<mainMenu>
    <menuBar>
        <subMenu>
            <label>Main</label>
            <subMenu>
                <label>Sub Menu</label>
                <subMenu>
                    <label>Sub-Sub Menu</label>
                    <scriptItem>
                        <label>Email</label>
                    </scriptItem>
                </subMenu>
                <separatorItem/> <!-- INSERTED !-->
                <scriptItem>
                    <label>Manager</label>
                </scriptItem>
            </subMenu>
            <subMenu>
                <label>Dev</label>
                <scriptItem>
                    <label>Dialog</label>
                </scriptItem>
            </subMenu>
            <separatorItem/> <!-- INSERTED !-->
            <scriptItem>
                <label>Sample</label>
            </scriptItem>
            <scriptItem>
                <label>Home</label>
            </scriptItem>
            <scriptItem>
                <label>Browser</label>
            </scriptItem>
            <scriptItem>
                <label>Check</label>
            </scriptItem>
            <scriptItem>
                <label>Open</label>
            </scriptItem>
            <scriptItem>
                <label>Close</label>
            </scriptItem>
        </subMenu>
    </menuBar>
</mainMenu>

0 回复 | 直到 2 年前

Jack Fleeting 2 年前

我会使用lxml,因为它对xpath的支持更好:

from lxml import etree as etree
menus = """[your xml above]"""

doc = etree.fromstring(menus.encode())
targets = doc.xpath('//subMenu[not(.//subMenu)][./scriptItem]')
for target in targets:
    sep = etree.fromstring("""<separatorItem/>""".encode())
    target.addnext(sep)       

print(etree.tostring(doc).decode())

输出应该是你想要的。