使用ElementTree提取inter标签name属性并按规则拼接组名

需求说明 #

• 提取XML里<inter>标签的name属性值
• 当该<inter>下符合run="jbs"且kind="obopay"的<work>标签带有groups属性时，把每个组名和name拼接成name-组名的格式
• 若没有groups属性，直接保留name

输入XML #

<abtshop>
    <dDirectory>dub</dDirectory>
    <S>statusd</S>
    <work>worklogs</work>
    <custs>
        <cust>nim-us</cust>
    </custs>

    <mileage>999</mileage>

    <defaults>
        <default type="mercley">
            <user>dairy</user>
            <exec>slm.sh</exec>
            <env>
                <var name="SAN_HOME">youyou-11</var>
            </env>
        </default>
    </defaults>
    <inters>
        <inter name="nim_turk" first-day="20230301" historical="20220103" market="multi">
            <works>
                <work kind="obopay" run="jbs">
                    <args>
                        <arg name="distance">180000</arg>
                    </args>
                </work>
                <work kind="silkb" run="jbs">
                    <args>
                        <arg name="distance">180000</arg>
                    </args>
                </work>
            </works>
        </inter>
        <inter name="nim_us_m" first-day="20230301" historical="20220103" market="lone">
            <works>
                <work kind="obopay" run="jbs" groups="groupA,groupB">
                    <args>
                        <arg name="distance">120000</arg>
                        <arg name="jbsopt">xmas_size=1200000</arg>
                        <arg name="jbsopt">of_obopaying_threads=2</arg>
                    </args>
                </work>
                <work kind="silkb" run="jbs" groups="groupA,groupB">
                    <args>
                        <arg name="distance">120000</arg>
                        <arg name="jbsopt">xmas_size=1200000</arg>
                    </args>
                </work>
            </works>
        </inter>
    </inters>
</abtshop>

尝试的错误代码 #

import xml.etree.ElementTree as ET

tree=ET.parse('test.xml')
root = tree.getroot()
xm_subs=[]
for subn in root.findall(".//inter/works/work[@run='jbs'][@kind='obopay']/../.."):
        sname=subn.attrib["name"]
        for subg in root.findall(".//inter[@name='%s']/jobs/job[@run='jbs'][@kind='obopay'][@groups]" % sname):
                        groups=subg.attrib['groups']
                        for gname in groups.split(","):
                                sub_name=subn.attrib["name"] + "-" + gname
                                xm_subs.append(sub_name)

        else:
                xm_subs.append(subn.attrib["name"])
print(xm_subs)

问题点分析 #

1. XPath路径错误：原代码里写了/jobs/job，但XML中对应的节点是/works/work，导致找不到目标节点
2. 循环逻辑错误：else和外层for绑定，而非判断是否找到带groups的节点，导致即使找到groups，也会额外添加原name值
3. 冗余查找：不需要从root重新定位节点，直接从当前inter节点内部查找更高效

修正后的代码 #

import xml.etree.ElementTree as ET

tree = ET.parse('test.xml')
root = tree.getroot()
xm_subs = []

# 遍历所有符合条件的inter节点
for inter in root.findall(".//inter[works/work[@run='jbs'][@kind='obopay']]"):
    inter_name = inter.attrib["name"]
    # 查找当前inter下符合条件且带groups属性的work节点
    work_node = inter.find(".//work[@run='jbs'][@kind='obopay'][@groups]")
    
    if work_node is not None:
        # 拆分组名并拼接
        groups = work_node.attrib["groups"].split(",")
        for group in groups:
            xm_subs.append(f"{inter_name}-{group}")
    else:
        # 没有groups属性，直接添加name
        xm_subs.append(inter_name)

print(xm_subs)

预期输出 #

['nim_turk', 'nim_us_m-groupA', 'nim_us_m-groupB']

内容的提问来源于stack exchange，提问作者iamsage