python lxml-修改属性 [英] python lxml - modify attributes

问题描述

from lxml import objectify, etree

root = etree.fromstring('''<?xml version="1.0" encoding="ISO-8859-1" ?>
<scenario>
<init>
    <send channel="channel-Gy">
        <command name="CER">
            <avp name="Origin-Host" value="router1dev"></avp>
            <avp name="Origin-Realm" value="realm.dev"></avp>
            <avp name="Host-IP-Address" value="0x00010a248921"></avp>
            <avp name="Vendor-Id" value="11"></avp>
            <avp name="Product-Name" value="HP Ro Interface"></avp>
            <avp name="Origin-State-Id" value="1094807040"></avp>
            <avp name="Supported-Vendor-Id" value="10415"></avp>
            <avp name="Auth-Application-Id" value="4"></avp>
            <avp name="Acct-Application-Id" value="0"></avp>
            <avp name="Vendor-Specific-Application-Id">
                <avp name="Vendor-Id" value="11"></avp>
                <avp name="Auth-Application-Id" value="4"></avp>
                <avp name="Acct-Application-Id" value="0"></avp>
            </avp>
            <avp name="Firmware-Revision" value="1"> </avp>
        </command>
    </send>
</init>

<traffic>
    <send channel="channel-Gy" >
        <action>
            <inc-counter name="HbH-counter"></inc-counter>
            ....
        </action>
    </send>
</traffic>
</scenario>''')

如何修改/设置两个值?

How can I modify/set both values?

• 主机IP地址 value ="0x00010a248921"

供应商ID" value ="11"

我尝试访问失败

root.xpath("//scenario/init/send_channel/command[@name='CER']/avp[@name='Host-IP-Address']/value/text()")

目标:我希望看到一个lxml.objectify与Xpath解决方案，但是我将接受其他基于lxml的解决方案.

Goal: I'd preferably like to see a lxml.objectify vs an Xpath solution but I'll accept other lxml based solutions.

文件< 100kB，因此速度/RAM无关紧要.

The files are <100kB so speed/RAM is not much of a concern.

推荐答案

import lxml.etree as et

tree = et.fromstring('''
... your xml ...
''')

for host_ip in tree.xpath("/scenario/init/send/command[@name='CER']/avp[@name='Host-IP-Address']"):
    host_ip.attrib['value'] = 'foo'

print et.tostring(tree)

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