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Python使用名稱空間解析Xml

[英]Python parsing the Xml with namespace

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>

這是我用來解析XMl以獲得“ param”和“ value”標簽的Python腳本。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,'CLIAggregation')(SM_HOUR_FORMAT,'24_HR')(MCNRecordsOrder,'LIFO')

您可以執行以下幾項操作來確保找到正確的文件-

我看不到以下行中提到的.XML文件的名稱-

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

確保在此代碼中輸入文件名稱,如下所示:

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

如果仍然無法使用,請嘗試定義文件的正確路徑-

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

這應該工作。 讓我知道是否有幫助。 :)

嘗試這個:

導入xml.etree.ElementTree作為ET導入系統

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)

這將給您想要的結果。

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