Home/ Questions/Is it okay to fix the typeerror: cannot use a string pattern on a bytes-like object problem?
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Colton Li
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Asked: 2022-05-17T14:19:13+00:00 In:

Is it okay to fix the typeerror: cannot use a string pattern on a bytes-like object problem?

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Here is the program I run:

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)

After I run, it returns an error:

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

Does anyone have any suggestions for the problem below: typeerror: cannot use a string pattern on a bytes-like object in the python – How to correct it?

cannot use a string pattern on a bytes-like object
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2 Answers

  1. Best Answer
    Expert

    The cause: The cause of the “typeerror: cannot use a string pattern on a bytes-like object in the python” is simply because your regex is a string, however html is made up of bytes:

    Solution: 

    You need to use .decode to convert html (a byte-like object) to a string.

    For example, this:

    html = response. read(). decode('utf-8').
    • 17

  2. Your regex is a string but html bytes. 

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

    Because python does not know the encoded bytes, it throws an error when you attempt to use string regex.

    You can decode add bytes to a string. 

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

    You can also use a bytes regex 

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

    This particular context allows you to get the encoding using the response headers. 

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

    For more information, see the urlopen documentation .

    • 7

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