Python使用名称空间解析Xml

时间:2019-05-23 09:47:44

标签: python xml.etree

需要解析的Xml “ cos1.XML”

<config xmlns="http://tail-f.com/ns/config/1.0">
  <sys xmlns="urn:XYZ:ns:yang:app:4.3.3.0">
  <app>
  <Feature>
    <name>0</name>
    <FeatureID>default</FeatureID>
    <param>MaxVoiceMessageLength</param>
    <value>120s</value>
  </Feature>
  <Feature>
    <name>96</name>
    <FeatureID>default</FeatureID>
    <param>MCNType</param>
    <value>CLIAggregation</value>
  </Feature>
  <Feature>
    <name>97</name>
    <FeatureID>default</FeatureID>
    <param>SM_HOUR_FORMAT</param>
    <value>24_HR</value>
  </Feature>
  <Feature>
    <name>99</name>
    <FeatureID>default</FeatureID>
    <param>MCNRecordsOrder</param>
    <value>LIFO</value>
  </Feature>
  </app>
  </sys>
</config>

这是我用来解析XMl以获得“ param”和“ value”标签的Python脚本。但是findall返回空。

import xml.etree.ElementTree as ET
import sys
def modifycos():

    tree = ET.parse(cos1.xml)
    root = tree.getroot()
    for cos in root.findall('./config/sys/app/Feature')
        parameter = cos.find('param').text
        parmvalue = cos.get('value')
        print(parameter, parmvalue)

modifycos()

(MaxVoiceMessageLength,'120s') (MCNType,“ CLIA聚合”) (SM_HOUR_FORMAT,“ 24_HR”) (MCNRecordsOrder,'LIFO')

2 个答案:

答案 0 :(得分:1)

您可以执行以下几项操作,以确保找到正确的文件-

我看不到以下行中提到的.XML文件的名称-

for cos in root.findall('./config/sys/app/Feature'):

请确保在此代码中输入文件名称,例如-

for cos in root.findall('./config/sys/app/Feature/cos1.XML'):

如果仍然无法正常运行,请尝试定义文件的正确路径-

import os
current_path = os.path.dirname(os.path.realpath(__file__))
file_path = os.path.join(current_path+'/config/sys/app/Feature/cos1.XML')

这应该有效。让我知道是否有帮助。 :)

答案 1 :(得分:0)

尝试一下:

将xml.etree.ElementTree导入为ET 导入系统

def modifycos():

    tree = ET.parse("try.xml")

    root = tree.getroot()

    sys = root.getchildren()[0]
    app = sys.getchildren()[0]
    features = app.getchildren()
    for element in features:
        childs = element.getchildren()
        for child in childs:
            if "param" in child.tag:
                parameter = child.text

            if "value" in child.tag:
                paramvalue = child.text
        print(parameter , paramvalue)

这将为您提供理想的结果。