操作系统模块 在Python中,提供了与操作系统交互的函数。OS属于Python的标准实用程序模块。该模块提供了一种使用操作系统相关功能的便携方式。
null
os.access()
方法使用真实的uid/gid来测试对路径的访问。大多数操作都使用有效的uid/gid,因此,可以在suid/sgid环境中使用此例程来测试调用用户是否具有指定的路径访问权限。
语法:
os.access(path, mode)
参数:
路径: 要测试访问或存在的路径 模式: 应该是F_OK以测试路径的存在,或者可以是R_OK、W_OK和X_OK中的一个或多个的包含或,以测试权限。
以下值可以作为access()的模式参数传递,以测试以下内容:
返回: 如果允许访问,则返回True,否则返回False。
代码#1: 了解access()方法
# Python program tyring to access # file with different mode parameter # importing all necessary libraries import os import sys # Different mode parameters will # return True if access is allowed, # else returns False. # Assuming only read operation is allowed on file # Checking access with os.F_OK path1 = os.access( "gfg.txt" , os.F_OK) print ( "Exists the path:" , path1) # Checking access with os.R_OK path2 = os.access( "gfg.txt" , os.R_OK) print ( "Access to read the file:" , path2) # Checking access with os.W_OK path3 = os.access( "gfg.txt" , os.W_OK) print ( "Access to write the file:" , path3) # Checking access with os.X_OK path4 = os.access( "gfg.txt" , os.X_OK) print ( "Check if path can be executed:" , path4) |
输出:
Exists the path: True Access to read the file: True Access to write the file: False Check if path can be executed: False
代码#2: 验证访问权限后打开文件的代码
# Python program to open a file # after validating the access # checking readability of the path if os.access( "gfg.txt" , os.R_OK): # open txt file as file with open ( "gfg.txt" ) as file : return file .read() # in case can't access the file return "Facing some issue" |
输出:
Facing some issue
© 版权声明
文章版权归作者所有,未经允许请勿转载。
THE END