Python 路径拼接函数os.path.join()

os.path.join()
功能：连接两个或更多的路径名组件

demo1 : 各组件名首字母不包含’/’

import os
path1 = 'aaa'
path2 = 'bbb'
path3 = 'ccc'
path10 = path1 + path2 + path3
path20 = os.path.join(path1,path2,path3)
print('path10 = ',path10)
print('path20 = ',path20)

path10 = aaabbbccc
path20 = aaa\bbb\ccc

demo2：有一个组件是绝对路径

import os
path1 = '/aaa'
path2 = 'bbb'
path3 = 'ccc'
path10 = path1 + path2 + path3
path20 = os.path.join(path1,path2,path3)
print('path10 = ',path10)
print('path20 = ',path20)

path10 = /aaabbbccc
path20 = /aaa\bbb\ccc

import os
path1 = '/aaa'
path2 = '/bbb'
path3 = 'ccc'
path10 = path1 + path2 + path3
path20 = os.path.join(path1,path2,path3)
print('path10 = ',path10)
print('path20 = ',path20)

path10 = /aaa/bbbccc
path20 = /bbb\ccc

path1 = '/aaa'
path2 = 'bbb'
path3 = '/ccc'
path10 = path1 + path2 + path3
path20 = os.path.join(path1,path2,path3)
print('path10 = ',path10)
print('path20 = ',path20)

path10 = /aaabbb/ccc
path20 = /ccc

demo3：有一个组建为‘ ./ ’

path1 = '/aaa'
path2 = './bbb'
path3 = 'ccc'
path10 = path1 + path2 + path3
path20 = os.path.join(path1,path2,path3)
print('path10 = ',path10)
print('path20 = ',path20)

path10 = /aaa./bbbccc
path20 = /aaa./bbb\ccc

path1 = 'aaa'
path2 = './bbb'
path3 = '/ccc'
path10 = path1 + path2 + path3
path20 = os.path.join(path1,path2,path3)
print('path10 = ',path10)
print('path20 = ',path20)

path10 = aaa./bbb/ccc
path20 = /ccc

demo4： 最后一个组件为空

path1 = 'aaa'
path2 = './bbb'
path3 = ' '
path10 = path1 + path2 + path3
path20 = os.path.join(path1,path2,path3)
print('path10 = ',path10)
print('path20 = ',path20)

path10 = aaa./bbb
path20 = aaa./bbb\

总结：

1. 该方法会从最后一个字符串开始向前寻找，直到找到以’ / ’ 开头的字符串，并将他们拼接起来
2. 如果各组件名首字母不包含’/’，则函数会自动加上
3. 如果有一个组件是绝对路径，则在它之前的所有组件均会被舍弃
4. 如果最后一个组件为空，则生成的路径以一个’\’分隔符结尾
5. 如果遇到‘ ./ ’ ，会把’ ./ ‘前面的字符一起拼接起来。
6. ’ / ’ 的优先级高于’ ./ ’ 。

参考博客：https://blog.youkuaiyun.com/qq_33468857/article/details/89027013
https://blog.youkuaiyun.com/weixin_40624269/article/details/82740282