python3.5 urllib.request.urlopen报错

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本文介绍了python3.5 urllib.request.urlopen报错的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

问 题

小白想自学python爬取网页信息,但是代码写了第二行就老是报错,度娘了很久感觉不是很能get,求大神点拨!

import urllib.request

Target = "http://www.tmsf.com/newhouse/property_330181_287055905_price.htm"
url = urllib.request.urlopen(Target)-----报错行
page = url.read()
url.close()
fp = open("grab.txt","wb")
fp.write(page)
fp.close()

报错信息
Traceback (most recent call last):
File "C:UsersjqhDesktop1.py", line 35, in <module>

url = urllib.request.urlopen(Target)

File "C:UsersjqhAppDataLocalProgramsPythonPython35liburllibrequest.py", line 162, in urlopen

return opener.open(url, data, timeout)

File "C:UsersjqhAppDataLocalProgramsPythonPython35liburllibrequest.py", line 471, in open

response = meth(req, response)

File "C:UsersjqhAppDataLocalProgramsPythonPython35liburllibrequest.py", line 581, in http_response

'http', request, response, code, msg, hdrs)

File "C:UsersjqhAppDataLocalProgramsPythonPython35liburllibrequest.py", line 509, in error

return self._call_chain(*args)

File "C:UsersjqhAppDataLocalProgramsPythonPython35liburllibrequest.py", line 443, in _call_chain

result = func(*args)

File "C:UsersjqhAppDataLocalProgramsPythonPython35liburllibrequest.py", line 589, in http_error_default

raise HTTPError(req.full_url, code, msg, hdrs, fp)

urllib.error.HTTPError: HTTP Error 403: Forbidden

解决方案

http 403 是对方网站返回的信息,你的代码没问题。

我用你的代码在python3.6.1下作了测试,可以正常运行,页面内容保存在grab.txt中。

这篇关于python3.5 urllib.request.urlopen报错的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

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