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Python Can';html2text是否不使用read()?_Python_Python 3.x_Urllib - Fatal编程技术网

Python Can';html2text是否不使用read()?

Python Can';html2text是否不使用read()?,python,python-3.x,urllib,Python,Python 3.x,Urllib,我正在制作一个Python程序,在网页上搜索一个单词。虽然,当我尝试 website = urllib.request.urlopen(url) content = website.read() website.close() test = html2text.html2text(content) print(test) 我得到这个错误: test = html2text.html2text(content) File "/Library/Frameworks/Python.framework

我正在制作一个Python程序,在网页上搜索一个单词。虽然,当我尝试

website = urllib.request.urlopen(url)
content = website.read()
website.close()
test = html2text.html2text(content)
print(test)
我得到这个错误:

test = html2text.html2text(content)
File "/Library/Frameworks/Python.framework/Versions/3.5/lib/python3.5/site-    packages/html2text/__init__.py", line 840, in html2text
return h.handle(html)
File "/Library/Frameworks/Python.framework/Versions/3.5/lib/python3.5/site-  packages/html2text/__init__.py", line 129, in handle
self.feed(data)
File "/Library/Frameworks/Python.framework/Versions/3.5/lib/python3.5/site-packages/html2text/__init__.py", line 125, in feed
data = data.replace("</' + 'script>", "</ignore>")
TypeError: a bytes-like object is required, not 'str'
test=html2text.html2text(内容)
html2text中的文件“/Library/Frameworks/Python.framework/Versions/3.5/lib/python3.5/site-packages/html2text/__init__.py”,第840行
返回h.handle(html)
文件“/Library/Frameworks/Python.framework/Versions/3.5/lib/python3.5/site-packages/html2text/_init_____;.py”,第129行,在handle中
self.feed(数据)
提要中的文件“/Library/Frameworks/Python.framework/Versions/3.5/lib/python3.5/site packages/html2text/_init___.py”,第125行
数据=数据。替换(“,”)
TypeError:需要类似字节的对象,而不是“str”
我是Python新手,因此不确定如何处理此错误。
Python 3.5,Mac。
decode()
字符集
头()中发送字符集的内容:


适用于我(Python3.5,Mac OS)。

我遇到了这个错误:
charset=resource.headers.get\u content\u charset()AttributeError:module'resource'没有属性'headers'
@Kamdroid您确定要在Python3.5上尝试吗?上面写着
Python2.7.10
,很奇怪。不过我确实下载了Python3。也许是因为在下载v3之前我下载了v2?虽然我在下载v3之前删除了Python2文件夹。但是我有一个名为python3.5的Python文件夹。@Kamdroid你试过以
python3.5
的形式运行它吗?嗯?我想我只有Python 3.5应用程序。Python2的目录是什么?
resource = urllib.request.urlopen(url)
content = resource.read()
charset = resource.headers.get_content_charset()
content = content.decode(charset)