用python编辑XML

Question

我正试图解析一个xml文件，在这个文件中我想获取objlocation的字符串并更改字符串的内容。在

以下是我的xml文件的内容：

<?xml version="1.0" encoding="UTF-8" standalone="yes"?>
<publish show="STATE">

    <pubgroup objtype="ELE" location="/user_data/STATE/ITEM/character/ANM/ANM_rig_WALK_sg_v001/ANM_rig_WALK_sg_v001.xml">

        <member objidx="15283942" objlabel="anm" objlocation="/user_data/STATE/ITEM/character/ANM/ANM_rig_WALK_sg_v001/ANM_rig_WALK_sg_v001.anm"/>

        <member objidx="15283952" objlabel="fbx" objlocation="/user_data/STATE/ITEM/character/ANM/ANM_rig_WALK_sg_v001/ANM_rig_WALK_sg_v001_M_WALK_None.fbx"/>

        <member objidx="15283962" objlabel="mov" objlocation="/user_data/STATE/ITEM/character/ANM/ANM_rig_WALK_sg_v001/ANM_rig_WALK_sg_v001.mov"/>

        <member objidx="15283972" objlabel="libraryinfo" objlocation="/user_data/STATE/ITEM/character/ANM/ANM_rig_WALK_sg_v001/ANM_rig_WALK_sg_v001.json"/>

        <member objidx="15283982" objlabel="thumbnail" objlocation="/user_data/STATE/ITEM/character/ANM/ANM_rig_WALK_sg_v001/ANM_rig_WALK_sg_v001.mng"/>

    </pubgroup>
</publish>

我尝试了.firstChild或{}，它将输出的内容作为我的xml文件打印出来。这是我试图解析的xml文件的列表，其格式大致相同。在

我就是想这么做

Answer 1

您可以使用ElementTree API轻松修改xml文件

from xml.etree.ElementTree import parse
doc = parse('data.xml')
root = doc.getroot()
for t in root.iterfind('pubgroup/member'):
    t.attrib['objlocation'] = "spam"

doc.write('output.xml', xml_declaration=True)

^{}方法返回一个generator函数，而不是list函数，如果您的xml文件非常大，这非常方便

输出

^{pr2}$

这里spam是objlocation新值。在

Answer 2

我能建议的最短代码：

from xml.etree.ElementTree import ElementTree
tree = ElementTree()
root = tree.parse('test.txt') # root represents <publish> tag

for member in root.findall('pubgroup/member'):
    print member.attrib['objlocation']

输出：

^{pr2}$

要进行更改：

for member in root.findall('pubgroup/member'):
    member.attrib['objlocation'] = 'changed'
tree.write('output.txt')