Python3不需要uuid:导入/usr/local/lib/python3.x/site-packages/uuid.py提示if not 0 time_low 32L错误

简介: Python3不需要uuid:导入/usr/local/lib/python3.x/site-packages/uuid.py提示if not 0 time_low 32L错误

这错误...

Traceback (most recent call last):
  File "/home/aaa.py", line 17, in <module>
    import requests
  File "/usr/local/lib/python3.5/site-packages/requests/__init__.py", line 53, in <module>
    from .packages.urllib3.contrib import pyopenssl
  File "/usr/local/lib/python3.5/site-packages/requests/packages/__init__.py", line 27, in <module>
    from . import urllib3
  File "/usr/local/lib/python3.5/site-packages/requests/packages/urllib3/__init__.py", line 8, in <module>
    from .connectionpool import (
  File "/usr/local/lib/python3.5/site-packages/requests/packages/urllib3/connectionpool.py", line 41, in <module>
    from .request import RequestMethods
  File "/usr/local/lib/python3.5/site-packages/requests/packages/urllib3/request.py", line 7, in <module>
    from .filepost import encode_multipart_formdata
  File "/usr/local/lib/python3.5/site-packages/requests/packages/urllib3/filepost.py", line 4, in <module>
    from uuid import uuid4
  File "/usr/local/lib/python3.5/site-packages/uuid.py", line 138
    if not 0 <= time_low < 1<<32L:
                                ^
SyntaxError: invalid syntax

怎么办?网上搜索了半天没找到办法,后来干脆卸载,问题解决!

sudo pip3 uninstall uuid
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