使用 bs4 提取 html 文件中的文本

Question

想从我的 html 文件中提取文本。如果我在下面使用特定文件：

import bs4, sys
from urllib import urlopen
#filin = open(sys.argv[1], 'r')
filin = '/home/iykeln/Desktop/R_work/file1.html' 
webpage = urlopen(filin).read().decode('utf-8')
soup = bs4.BeautifulSoup(webpage)
for node in soup.findAll('html'):
    print u''.join(node.findAll(text=True)).encode('utf-8')

它会起作用。但是在下面尝试使用 open(sys.argv[1], 'r') 获取非特定文件：

import bs4, sys
from urllib import urlopen
filin = open(sys.argv[1], 'r')
#filin = '/home/iykeln/Desktop/R_work/file1.html' 
webpage = urlopen(filin).read().decode('utf-8')
soup = bs4.BeautifulSoup(webpage)
for node in soup.findAll('html'):
    print u''.join(node.findAll(text=True)).encode('utf-8')

或者

import bs4, sys
from urllib import urlopen
with open(sys.argv[1], 'r') as filin:
    webpage = urlopen(filin).read().decode('utf-8')
    soup = bs4.BeautifulSoup(webpage)
    for node in soup.findAll('html'):
        print u''.join(node.findAll(text=True)).encode('utf-8')

我将收到以下错误：

Traceback (most recent call last):
  File "/home/iykeln/Desktop/py/clean.py", line 5, in <module>
    webpage = urlopen(filin).read().decode('utf-8')
  File "/usr/lib/python2.7/urllib.py", line 87, in urlopen
    return opener.open(url)
  File "/usr/lib/python2.7/urllib.py", line 180, in open
    fullurl = unwrap(toBytes(fullurl))
  File "/usr/lib/python2.7/urllib.py", line 1057, in unwrap
    url = url.strip()
AttributeError: 'file' object has no attribute 'strip'

Answer 1

您不应该调用open，只需将文件名传递给urlopen：

import bs4, sys
from urllib import urlopen

webpage = urlopen(sys.argv[1]).read().decode('utf-8')
soup = bs4.BeautifulSoup(webpage)
for node in soup.findAll('html'):
    print u''.join(node.findAll(text=True)).encode('utf-8')

仅供参考，您不需要urllib打开本地文件：

import bs4, sys

with open(sys.argv[1], 'r') as f:
    webpage = f.read().decode('utf-8')

soup = bs4.BeautifulSoup(webpage)
for node in soup.findAll('html'):
    print u''.join(node.findAll(text=True)).encode('utf-8')

希望有帮助。