Python 如何使用lxml查找所有src标记并替换它们
我想使用lxml获取src内容,并用空格替换它们。 但车身仍然无法更换 请帮帮我谢谢你Python 如何使用lxml查找所有src标记并替换它们,python,html,html-parsing,lxml,lxml.html,Python,Html,Html Parsing,Lxml,Lxml.html,我想使用lxml获取src内容,并用空格替换它们。 但车身仍然无法更换 请帮帮我谢谢你 import re import lxml.html #the content of source.log is a webpage source code I got by scrapy with open("source.log", "r") as bb: c_str = bb.read() body = c_str.decode('utf-8') doc = lxml.html.
import re
import lxml.html
#the content of source.log is a webpage source code I got by scrapy
with open("source.log", "r") as bb:
c_str = bb.read()
body = c_str.decode('utf-8')
doc = lxml.html.fromstring(body)
src = doc.xpath("//@src")
for ss in src:
re.search(ss,body)
body.replace(str(ss),'')
print body
例如:
如果尸体是
'src="http://pic/1379181836.jpg"/><br>紅心<br></div><div>tel:12345678</div>' \
'src="http://pic/4447918.jpg"/>'
我想要的结果是:
'src=""/><br>紅心<br></div><div>tel:12345678</div>' \
'src=""/>'
至少,您需要将替换的结果指定给主体: 不过,我个人不喜欢这种方法。最好找到所有具有src属性的标记,并将属性值设置为空字符串:
for element in doc.xpath("//*[@src]"):
element.attrib['src'] = ''
print lxml.html.tostring(doc)
非常感谢。你是对的。你的代码非常漂亮。
for element in doc.xpath("//*[@src]"):
element.attrib['src'] = ''
print lxml.html.tostring(doc)