python提取xml里面的链接源码详解

 #coding=utf-8 import urllib import urllib.request import re url=&＃039;http://zhimo.yuanzhumuban.cc/sitemaps.xml&＃039; html=urllib.request.urlopen(url).read() html=html.decode(&＃039;utf-8&＃039;) r=re.compile(r&＃039;(http://zhimo.yuanzhumuban.cc.*?\.html)&＃039;) big=re.findall(r,html) for i in big: print(i) op_xml_txt=open(&＃039;xml.txt&＃039;,&＃039;a&＃039;) op_xml_txt.write(&＃039;%s\n&＃039;%i)

 import xml.dom.minidom def find_child(Par_nodes, mystr): for child_node in Par_nodes: if(len(child_node.childNodes) > 0): mystr = find_child(child_node.childNodes, mystr) elif(child_node.nodeValue != None): mystr += child_node.data.replace(&＃039;\n&＃039;, &＃039;&＃039;) return mystr if __name__ == &＃039;__main__&＃039;: dom1 = xml.dom.minidom.parse(&＃039;2.XML&＃039;) #打开xml文件 root = dom1.documentElement     #得到文档元素对象 app_nums = root.getEl来源gao@.dai.ma.com搞@^&[email protected])网ementsByTagName(&＃039;base:DocNumber&＃039;) #按标签名称查找，返回标签结点数组 app_num = app_nums[2] print(&＃039;专利申请号：&＃039;+app_num.firstChild.data) titles = root.getElementsByTagName(&＃039;business:InventionTitle&＃039;) title = titles[0] print(&＃039;专利名称：&＃039;+title.firstChild.data) Paragraphs = root.getElementsByTagName(&＃039;base:Paragraphs&＃039;) abstract = Paragraphs[0] print(&＃039;专利摘要：&＃039;+abstract.firstChild.data) company_names = root.getElementsByTagName(&＃039;base:Name&＃039;) company_name = company_names[0] print(&＃039;公司名称：&＃039;+company_name.firstChild.data) mystr = &＃039;&＃039; for i in range(len(Paragraphs)): if (Paragraphs[i].firstChild.data == &＃039;发明内容\n\t&＃039;): i+=1 while Paragraphs[i].firstChild.data != &＃039;附图说明\n\t&＃039;: mystr = find_child(Paragraphs[i].childNodes, mystr) i+=1 print(&＃039;发明内容：&＃039; + mystr)