monitor_module.serials_in_use.remove(serial_port)# Remove cachekeep_cache = Preferences().get('keep_cache',True)if(keep_cache):returnfile_path = Tools.getPathFromView(view)if(notfile_path):returnfile_name = Tools.getFileNameFromPath(file_path, ext=False) tmp_path = Paths.getTempPath()...
通过使用Path对象,可以更方便地获取路径下的文件名。下面是一个示例代码: frompathlibimportPathdefget_file_names(path):""" 获取指定路径下的所有文件名 """file_names=[str(file.name)forfileinPath(path).iterdir()iffile.is_file()]returnfile_names# 调用函数并打印结果path="/path/to/directory"names=g...
But if you wanted to get just the file name, how would you go about that? It took me a little while to find an answer, and the method not super obvious, so I’ll post it here. importglob,osfilePaths =glob.glob("C:\\Temp\\*.txt")forfilePathinfilePaths:printos.path.basename(fil...
# 需要导入模块: from src.tools.path import Path [as 别名]# 或者: from src.tools.path.Path importget_filename[as 别名]defcreate_chapter(self, src, title):template = self.get_template('directory','item_root') item = template.format(href=Path.get_filename(src), title=title)ifself.chapt...
下面是使用Path对象的示例代码: frompathlibimportPathdefget_file_path(filename):current_dir=Path.cwd()# 获取当前工作目录file_path=current_dir.joinpath(filename)# 拼接文件名和路径returnfile_path# 使用示例filename="example.txt"path=get_file_path(filename)print("文件路径:",path) ...
1. path()包装下, 路径拼接:FILE_ROOT.joinpath(r"Labels") 文件的创建LABELS_ROOT.mkdir(exist_ok=True), 文件list的遍历root.iterdir() '''test_path_val_list()''' 2. 取路径的尾巴和改变换格式txt'''getPathTail()''' 获取文件路径的方法: ...
Remove Filename From File Path Using the Pathlib Module in Python Conclusion File operations are a fundamental part of python application development. In this article, we will discuss how we can get the directory name from the file path in python. Later, we will also discuss how we can remov...
(__file__))[1]: ',os.path.split(os.path.realpath(__file__))[1])#其他文件引用本文件,调用该方法,获取执行文件的文件名称@staticmethoddefgetPerFileName():print('os.path.basename(sys.argv[0]: ',os.path.basename(sys.argv[0]))#获取本文件目录名称(不管执行位置)@staticmethoddefgetFile...
path = "/home/user/Documents/example.txt" folder_names = get_folder_names(path) print("每个文件夹的名称:", folder_names) 运行以上代码,输出结果为: 每个文件夹的名称: ['home', 'user', 'Documents'] 通过定义get_folder_names()函数,我们可以方便地获取路径字符串中每个文件夹的名称,并以列表的形...
import os from win32com.shell import shell def getDocPath(pathID=5): '''默认返回我的文档路径,出错则返回当前工作路径''' try: return shell.SHGetFolderPath(0, pathID, None, 0) except: return os.path.dirname(os.path.realpath(__file__)) print(getDocPath()) print(getDocPath(60))文章标...