from mechanize import Browser
br = Browser()
br.open('http://somewebpage')
html = br.response().readlines()
for line in html:
print line
当在HTML文件中打印一行时，我试图找到一种方法，只显示每个HTML元素的内容，而不是格式本身。如果它发现'<a href="等等。例如">some text</a>'，它只会打印'some text'， '<b>hello</b>'打印'hello'，等等。该怎么做呢?
当前回答
我正在解析Github自述，我发现下面的工作真的很好:
import re
import lxml.html
def strip_markdown(x):
links_sub = re.sub(r'\[(.+)\]\([^\)]+\)', r'\1', x)
bold_sub = re.sub(r'\*\*([^*]+)\*\*', r'\1', links_sub)
emph_sub = re.sub(r'\*([^*]+)\*', r'\1', bold_sub)
return emph_sub
def strip_html(x):
return lxml.html.fromstring(x).text_content() if x else ''
然后
readme = """<img src="https://raw.githubusercontent.com/kootenpv/sky/master/resources/skylogo.png" />
sky is a web scraping framework, implemented with the latest python versions in mind (3.4+).
It uses the asynchronous `asyncio` framework, as well as many popular modules
and extensions.
Most importantly, it aims for **next generation** web crawling where machine intelligence
is used to speed up the development/maintainance/reliability of crawling.
It mainly does this by considering the user to be interested in content
from *domains*, not just a collection of *single pages*
([templating approach](#templating-approach))."""
strip_markdown(strip_html(readme))
正确移除所有markdown和html。
其他回答
如果你需要保留HTML实体(即&)，我在Eloff的答案中添加了“handle_entityref”方法。
from HTMLParser import HTMLParser
class MLStripper(HTMLParser):
def __init__(self):
self.reset()
self.fed = []
def handle_data(self, d):
self.fed.append(d)
def handle_entityref(self, name):
self.fed.append('&%s;' % name)
def get_data(self):
return ''.join(self.fed)
def html_to_text(html):
s = MLStripper()
s.feed(html)
return s.get_data()
这是一个快速修复，甚至可以更优化，但它将工作良好。这段代码将用""替换所有非空标记，并从给定的输入文本中剥离所有html标记。你可以使用./file.py输入输出运行它
#!/usr/bin/python
import sys
def replace(strng,replaceText):
rpl = 0
while rpl > -1:
rpl = strng.find(replaceText)
if rpl != -1:
strng = strng[0:rpl] + strng[rpl + len(replaceText):]
return strng
lessThanPos = -1
count = 0
listOf = []
try:
#write File
writeto = open(sys.argv[2],'w')
#read file and store it in list
f = open(sys.argv[1],'r')
for readLine in f.readlines():
listOf.append(readLine)
f.close()
#remove all tags
for line in listOf:
count = 0;
lessThanPos = -1
lineTemp = line
for char in lineTemp:
if char == "<":
lessThanPos = count
if char == ">":
if lessThanPos > -1:
if line[lessThanPos:count + 1] != '<>':
lineTemp = replace(lineTemp,line[lessThanPos:count + 1])
lessThanPos = -1
count = count + 1
lineTemp = lineTemp.replace("<","<")
lineTemp = lineTemp.replace(">",">")
writeto.write(lineTemp)
writeto.close()
print "Write To --- >" , sys.argv[2]
except:
print "Help: invalid arguments or exception"
print "Usage : ",sys.argv[0]," inputfile outputfile"
# This is a regex solution.
import re
def removeHtml(html):
if not html: return html
# Remove comments first
innerText = re.compile('<!--[\s\S]*?-->').sub('',html)
while innerText.find('>')>=0: # Loop through nested Tags
text = re.compile('<[^<>]+?>').sub('',innerText)
if text == innerText:
break
innerText = text
return innerText.strip()
下面是一个简单的解决方案，基于惊人的快速lxml库剥离HTML标签并解码HTML实体:
from lxml import html
def strip_html(s):
return str(html.fromstring(s).text_content())
strip_html('Ein <a href="">schöner</a> Text.') # Output: Ein schöner Text.
这是一个类似于目前接受的答案(https://stackoverflow.com/a/925630/95989)的解决方案，除了它直接使用内部HTMLParser类(即没有子类化)，从而使它显着更简洁:
def strip_html(text):
parts = []
parser = HTMLParser()
parser.handle_data = parts.append
parser.feed(text)
return ''.join(parts)
