python 操作xml

xml.etree.ElementTree模块（简称 ET）

1.加载模块

xml.etree.cElementTree的底层是c，所以速度快，如果没有就只能加载xml.etree.ElementTree了

try: 
  import xml.etree.cElementTree as ET 
except ImportError: 
  import xml.etree.ElementTree as ET

2.载入xml文档

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

3.获取根节点

root = tree.getroot()

4.查找节点

allobj=root.findall('country')
obj=root.find('country')
subobj=root.find('subcountry')

5.获取节点的值

subobjtext=subobj.text
subobjtext=root.find('subcountry').text

6.获取节点属性

name = country.get('name') #子节点下属性name的值

添加节点

from xml.etree.ElementTree import Element, SubElement
方法1
root = Element("root")
child = Element("child")
root.append(child)
方法2
root = Element("root")
child = SubElement(root, "child")

节点赋值

child.text="zifuchuan"

#!/usr/bin/evn python 
#coding:utf-8 
  
try: 
  import xml.etree.cElementTree as ET 
except ImportError: 
  import xml.etree.ElementTree as ET 
import sys 
  
try: 
  tree = ET.parse("country.xml")     #打开xml文档 
  #root = ET.fromstring(country_string) #从字符串传递xml 
  root = tree.getroot()         #获得root节点  
except Exception, e: 
  print "Error:cannot parse file:country.xml."
  sys.exit(1) 
print root.tag, "---", root.attrib  
for child in root: 
  print child.tag, "---", child.attrib 
  
print "*"*10
print root[0][1].text   #通过下标访问 
print root[0].tag, root[0].text 
print "*"*10
  
for country in root.findall('country'): #找到root节点下的所有country节点 
  rank = country.find('rank').text   #子节点下节点rank的值 
  name = country.get('name')      #子节点下属性name的值 
  print name, rank 
     
#修改xml文件 
for country in root.findall('country'): 
  rank = int(country.find('rank').text) 
  if rank > 50: 
    root.remove(country) 
  
tree.write('output.xml')

7.查找并修改某个值

def changeXmlsLabel(xmlFolder,currentLabels,aimLabels):
    '''
    修改标注结果的类型名字
    xmlFolder='/home/lfs/dateset/sdjw/train/Annotations'
    currentLabels=['OuterProtrusion','InnerProtrusion']
    aimLabels=['defect','defect']
    changeXmlsLabel(xmlFolder,currentLabels,aimLabels)
    '''
    import xml.etree.ElementTree as ET
    xmllist=os.listdir(xmlFolder)
    num=0
    for xmlname in xmllist:
        xmlpathname=os.path.join(xmlFolder,xmlname)
        tree = ET.parse(xmlpathname)
        root = tree.getroot()
        objects=root.findall('object')
        for obj in objects:
            for name in obj.iter('name'):
                id=currentLabels.index(name.text)
                clabeltext=aimLabels[id]
                name.text=clabeltext
        tree.write(xmlpathname,encoding='utf-8')
        num+=1
        print('num:%d,name:%s'%(num,xmlname))

构建新的xml文件

#生成根元素
root=ET.Element("root")
#生成子元素 A
a=ET.Element("A")
#增加元素A的子元素 child1
a_child=ET.SubElement(a,"child1")
a_child.text="I'm child of A"
#增加元素A的子元素 child2
a_child1=ET.SubElement(a,"child2")
#生成子元素B
b=ET.Element("B")
#增加元素B的子元素child1
b_child=ET.SubElement(b,"child1")
b_child.text="I'm child of B"
b_child.set("name","book") #set() 接收的是 key,value 形式

#将a和b 组成一个元组传入extend()方法中，元素 A和B作为根元素的子元素
root.extend((a,b))
trees=ET.ElementTree(root)
#将trees 写入到文件 test4.xml, 内容为 <root><A><child1>I'm child of A</child1><child2 /></A><B><child1 name="book">I'm child of B</child1></B></root>
trees.write(os.path.join(os.path.dirname(__file__),"test4.xml"))

解决排版缩进问题

def indent(elem, level=0):
    i = "\n" + level*"\t"
    if len(elem):
        if not elem.text or not elem.text.strip():
            elem.text = i + "\t"
        if not elem.tail or not elem.tail.strip():
            elem.tail = i
        for elem in elem:
            indent(elem, level+1)
        if not elem.tail or not elem.tail.strip():
            elem.tail = i
    else:
        if level and (not elem.tail or not elem.tail.strip()):
            elem.tail = i
indent(root)
trees=ET.ElementTree(root)
trees.write(os.path.join(os.path.dirname(__file__),"test4.xml"))

参考：用Python解析XML的几种常见方法的介绍

最后编辑于：2021.08.17 18:45:54