目录处理 OS
目录处理
目录-->路径，文件夹 文件：txt
1. 新建和删除一个目录
import os #引入os目录
from xx import xx
os.mkdir("D:\PycharmProjects\RobotFramework\vda_pakage\Learning\python9") #绝对路径
os.rmdir("python8") #相对路径
2. 获取目录 文件的路径 __file__指定的是当前的文件
print(os.path.realpath(__file__)) #返回绝对路径及该路径下当前的文件
print(os.getcwd()) #返回当前绝对路径
D:PycharmProjectsRobotFrameworkvda_pakageLearningclass_07_02.py
D:PycharmProjectsRobotFrameworkvda_pakageLearning
print(os.path.dirname(__file__)) #返回指定路径的目录
print(os.path.basename(__file__)) #返回当前工作的文件名
D:/PycharmProjects/RobotFramework/vda_pakage/Learning
class_07_02.py
3. 拼接路径以及拼接之后新建
path = "D:\PycharmProjects\RobotFramework\vda_pakage\Learning"
new_path = os.path.join(path, "python10")
print(new_path)
# D:PycharmProjectsRobotFrameworkvda_pakageLearningpython10
拼接之后再去新建目录，只能新建一级，不能跨级
os.mkdir(new_path)
new_path_1 = os.path.join(path, "python8\python9", "python10")
print(new_path_1)
os.mkdir(new_path_1) #D:PycharmProjectsRobotFrameworkvda_pakageLearningpython8python9python10
4. 获取目录信息，判断目录的性质
print(os.listdir("D:\PycharmProjects\RobotFramework\vda_pakage\Learning"))
os.listdir返回的结果是列表类型的数据,返回给出的路径下的文件和目录
['class1.py', 'class_07_02.py', 'hm_0630.py', 'path.py', 'python10', 'python8', 'python9', 'task2_06_25.py', 'task3_06_27.py', 'task3_addition_06_27.py', 'task4_06_30_fresh.py', 'task4_06_30_middle.py', 'task4_06_30_primary.py', '__init__.py']
判断文件或是目录的性质，返回的数据是布尔类型（True/False）
print(os.path.isfile(__file__)) #True
print(os.path.isdir(__file__)) #False
5. 对文件路径进行切割split,返回两个值，数据类型为元组
print(os.getcwd())
print(os.path.split(os.getcwd()))
D:PycharmProjectsRobotFrameworkvda_pakageLearning
('D:\PycharmProjects\RobotFramework\vda_pakage', 'Learning')
print(os.path.split(os.path.realpath(__file__)))
('D:\PycharmProjects\RobotFramework\vda_pakage\Learning', 'class_07_02.py')
#当前目录下创建3个文件夹，命名为test0，test1，test2
for i in range(3):
os.mkdir('test'+str(i))