Python之文件路径名的操作

使用 os.path 模块中的函数来操作路径名

import os

# 获取当前文件路径

path=os.path.abspath(__file__)  # 获取绝对路径 /home/zzy/PycharmProjects/MyTest/testPackage/secondPackage/文件与IO.py

print(path)

print(os.path.realpath(__file__))  # 获取真实路径  /home/zzy/PycharmProjects/MyTest/testPackage/secondPackage/文件与IO.py

# 文件基础名字

print(os.path.basename(path))  # 文件与IO.py

# 获取文件目录

print(os.path.dirname(path))   # /home/zzy/PycharmProjects/MyTest/testPackage/secondPackage

# 文件路径拼接

print(os.path.join("new","zzy",os.path.basename(path)))  # new/zzy/文件与IO.py

# 展开用户的主目录

print(os.path.expanduser("~/PycharmProjects/MyTest/testPackage/secondPackage"))   # /home/zzy/PycharmProjects/MyTest/testPackage/secondPackage

# 拆分文件扩展名

print(os.path.splitext(path))  # ('/home/zzy/PycharmProjects/MyTest/testPackage/secondPackage/文件与IO', '.py')

# 拆分文件名和目录

print(os.path.split(path))  # ('/home/zzy/PycharmProjects/MyTest/testPackage/secondPackage', '文件与IO.py')

对于任何的文件名的操作，你都应该使用 os.path 模块,特别是为了可移植性考虑的时候更应如此，
因为 os.path 模块知道Unix和Windows系统之间的差异并且能够可靠地处理类似

测试文件是否存在

print(os.path.exists("MyTest/testPackage"))  # False

print(os.path.exists("/home/zzy/PycharmProjects/MyTest/testPackage"))  # True

判断文件类型

print(os.path.isdir("/home/zzy/PycharmProjects/MyTest/testPackage/secondPackage"))  # True

print(os.path.isdir("/home/zzy/PycharmProjects/MyTest/testPackage/secondPackage/文件与IO.py"))  # False

print(os.path.isfile("/home/zzy/PycharmProjects/MyTest/testPackage/secondPackage/文件与IO.py"))  # True

print(os.path.islink("/home/zzy/PycharmProjects/MyTest/testPackage/secondPackage/文件与IO.py"))  # False

print(os.path.islink("/home/zzy/PycharmProjects/MyTest/testPackage/secondPackage"))  # False

巴特西

Python之文件路径名的操作

最新文章

热门文章