亲宝软件园·资讯

展开

Python 解析xml Python 解析xml文件的示例

南风丶轻语 人气:0
想了解Python 解析xml文件的示例的相关内容吗,南风丶轻语在本文为您仔细讲解Python 解析xml的相关知识和一些Code实例,欢迎阅读和指正,我们先划重点:python,XML,python,解析xml,下面大家一起来学习吧。

1、获取xml树

import xml.etree.ElementTree as ET


def getTree(xmlName):
  xmlName = xmlName.strip()
  try:
    tree = ET.parse(xmlName)
  except:
    tree = None
    print 'Analysis xml file fail,file name: {}'.format(xmlName)
  return tree

2、获取根节点

def getRoot(tree):
  if tree is not None:
    root = tree.getroot()
  else:
    root = None
    print 'Get root fail'
  return root

3、查看根节点

def seeRoot(root):
  '''<country name="tan">我是小明</country>'''
  if root is not None:
    print 'root tag:', root.tag # 标签(country)
    print 'root attrib:', root.attrib # 屬性(name="tan")
    print 'root text:', root.text # 文本(我是小明)
    print 'root tail:', root.tail # 尾字符串(未涉及)

4、从根开始遍历树

def traverseRoot(root):
  if root is not None:
    for label1 in root:
      print 'label1 tag:', label1.tag
      print 'label1 attrib:', label1.attrib
      print 'label1 text:', label1.text
      print 'label1 tail:', label1.tail
      print '=================='
      for label2 in label1:
        print 'label2 tag:', label2.tag
        print 'label2 attrib:', label2.attrib
        print 'label2 text:', label2.text
        print 'label2 tail:', label2.tail
        print '=================='
        for label3 in label2:
          print 'label3 tag:', label3.tag
          print 'label3 attrib:', label3.attrib
          print 'label3 text:', label3.text
          print 'label3 tail:', label3.tail
          print '=================='

5、找到2012年的gdppc和neighbor下的b标签(找到同层有条件的同层另一个tag的文本)

def findYouNedd(root):
  '''查找year为2012下的b标签的文本'''
  if root is not None:
    for label1 in root:
      for label2 in label1:
        if label1.tag == 'country' and label2.text == '2012': # 找到本层标签为country且下一层有2012文本
          print 'Find tag为country and next year=2012'
          for child in label1:
            if child.tag == 'gdppc':
              print child.text
            for youNeed in child:
              if youNeed.tag == 'b':
                print 'You need:', youNeed.text

6、查找父节点下的子节点

def findChildNode(fatherNode, childNode):
  childNode = childNode.strip()
  if fatherNode is not None:
    childs = fatherNode.findall(childNode)
    print childs
    print len(childs)

7、另一种办法实现第4点

def findYouNedd2(root):
  countryNodes = root.findall('country')
  if root is not None:
    for countryNode in countryNodes:
      if countryNode.find('year').text == '2012':
        print countryNode.find('gdppc').text

8、移除节点

def delNode(tree, nodeName):
  nodeName = nodeName.strip()
  if tree is not None:
    root = tree.getroot()
    findNode = root.find(nodeName)
    if findNode is not None and findNode.tag == nodeName:
      root.remove(findNode)
  tree.write('removeNode.xml') # 移除节点后新的xml

9、xml样例(xmlDemo.xml)

<?xml version="1.0"?>
<data>
  <country name="Liechtenstein">
    <rank>1</rank>
    <year>2008</year>
    <gdppc>141100</gdppc>
    <neighbor name="Austria" direction="E"/>
    <neighbor name="Switzerland" direction="W"/>
  </country>
  <country name="Singapore">
    <rank>4</rank>
    <year>2011</year>
    <gdppc>59900</gdppc>
    <neighbor name="Malaysia" direction="N">123
      <a name="a"> aaa </a>
    </neighbor>
  </country>
  <country name="Singapore">
    <rank>68</rank>
    <year>2012</year>
    <gdppc>13600</gdppc>
    <neighbor name="Costa Rica" direction="W"/>
    <neighbor name="Colombia" direction="E">456
      <b name="b"> bbb </b>
    </neighbor>
  </country>
  <city>789</city>
</data>

加载全部内容

相关教程
猜你喜欢
用户评论