TypeError:无法在re.findall()中的类似字节的对象上使用字符串模式


106

我正在尝试学习如何自动从页面获取URL。在下面的代码中,我试图获取网页的标题:

import urllib.request
import re

url = "http://www.google.com"
regex = r'<title>(,+?)</title>'
pattern  = re.compile(regex)

with urllib.request.urlopen(url) as response:
   html = response.read()

title = re.findall(pattern, html)
print(title)

我得到这个意外的错误:

Traceback (most recent call last):
  File "path\to\file\Crawler.py", line 11, in <module>
    title = re.findall(pattern, html)
  File "C:\Python33\lib\re.py", line 201, in findall
    return _compile(pattern, flags).findall(string)
TypeError: can't use a string pattern on a bytes-like object

我究竟做错了什么?


Answers:



28

问题是您的正则表达式是一个字符串,但是htmlbytes

>>> type(html)
<class 'bytes'>

由于python不知道这些字节是如何编码的,因此当您尝试在它们上使用字符串正则表达式时会引发异常。

您可以decode将字节存储为字符串:

html = html.decode('ISO-8859-1')  # encoding may vary!
title = re.findall(pattern, html)  # no more error

或使用字节正则表达式:

regex = rb'<title>(,+?)</title>'
#        ^

在此特定上下文中,您可以从响应标头中获取编码:

with urllib.request.urlopen(url) as response:
    encoding = response.info().get_param('charset', 'utf8')
    html = response.read().decode(encoding)

有关更多详细信息,请参见urlopen文档

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