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在python中解析网页时遇到错误_Python_Python 3.x_Python 2.7_Python Requests_Urllib - Fatal编程技术网

在python中解析网页时遇到错误

在python中解析网页时遇到错误,python,python-3.x,python-2.7,python-requests,urllib,Python,Python 3.x,Python 2.7,Python Requests,Urllib,我有一个python脚本,它以url作为输入,通过urllib打开url并解析网页。但我得到了以下错误: Traceback (most recent call last): File "C:\Users\Arjun Gaur\Desktop\MyPython\arjurl.py", line 1, in <module> import urllib.request, urllib.parse, urllib.error File "C:\Users\Arjun Ga

我有一个python脚本,它以url作为输入,通过urllib打开url并解析网页。但我得到了以下错误:

Traceback (most recent call last):
  File "C:\Users\Arjun Gaur\Desktop\MyPython\arjurl.py", line 1, in <module>
    import urllib.request, urllib.parse, urllib.error
  File "C:\Users\Arjun Gaur\Desktop\MyPython\urllib.py", line 1, in <module>
    import urllib.request
ModuleNotFoundError: No module named 'urllib.request'; 'urllib' is not a package

请在问题中包括回溯。使用回溯编辑似乎您有一个名为
urllib.py
的文件,您不小心导入了该文件。thnx,我在同一目录中有一个同名文件。但现在遇到了不同的错误您只是打印一条消息以供输入,而不是获取实际的用户输入。将
url=print('输入链接:')
更改为
url=input('输入链接:')
import urllib.request
import urllib.parse
import urllib.error
import BeautifulSoup
import ssl


ctx = ssl.create_default_context()
ctx.check_hostname = False
ctx.verify_mode = ssl.CERT_NONE

url = print('Enter Link : ')
urlstr = urllib.request.urlopen(url,context=ctx).read()

soup = BeautifulSoup(urlstr,'html-parser')

tags = soup('a')

for tag in tags :
    print(tag.get('href',None))