Python tkinter 之文件对话框(filedialog)

2023-12-13 13:59:04

1 文件

1.1 获取单个文件名称:askopenfilename()

import tkinter
from tkinter import filedialog


class FileDiaLogDemo(object):
    def __init__(self):
        window = tkinter.Tk()
        window.withdraw()  # 不显示窗体

    def get_file_name(self):
        # 选择单个文件(绝对路径)
        file_path = filedialog.askopenfilename()
        print(file_path)


if __name__ == '__main__':
    test = FileDiaLogDemo()
    test.get_file_name()

测试结果(参考):

C:/Users/Administrator/Desktop/Temp/1.txt

1.2 获取多个文件名称:askopenfilenames()

import tkinter
from tkinter import filedialog


class FileDiaLogDemo(object):
    def __init__(self):
        window = tkinter.Tk()
        window.withdraw()  # 不显示窗体

    def get_file_names(self):
        # 选择多个文件(绝对路径,Ctrl + 文件)
        file_path = filedialog.askopenfilenames()
        print(f'返回值类型:{type(file_path)}')  # <class 'tuple'>
        print(file_path)


if __name__ == '__main__':
    test = FileDiaLogDemo()
    test.get_file_names()

测试结果(参考):

('C:/Users/Administrator/Desktop/Temp/1.txt', 
 'C:/Users/Administrator/Desktop/Temp/2.txt')

1.3 获取单个文件属性:askopenfile()

import tkinter
from tkinter import filedialog


class FileDiaLogDemo(object):
    def __init__(self):
        window = tkinter.Tk()
        window.withdraw()  # 不显示窗体

    def get_file(self):
        # 选择单个文件
        file = filedialog.askopenfile()
        print(f'文件的名称:{file.name}')
        print(f'文件的模式:{file.mode}')
        print(f'文件的编码:{file.encoding}')


if __name__ == '__main__':
    test = FileDiaLogDemo()
    test.get_file()

1.4 获取多个文件属性:askopenfiles()

import tkinter
from tkinter import filedialog


class FileDiaLogDemo(object):
    def __init__(self):
        window = tkinter.Tk()
        window.withdraw()  # 不显示窗体

    def get_files(self):
        # 选择单个文件
        file = filedialog.askopenfiles()
        print(f'返回值类型:{type(file)}')  # <class 'list'>
        print(f'文件的个数:{len(file)}')


if __name__ == '__main__':
    test = FileDiaLogDemo()
    test.get_files()

1.5 获取保存文件的路径:asksaveasfilename()

import tkinter
from tkinter import filedialog


class FileDiaLogDemo(object):
    def __init__(self):
        window = tkinter.Tk()
        window.withdraw()  # 不显示窗体

    def get_save_file_name(self):
        file = filedialog.asksaveasfilename(title="请选择文件存储路径",
                                            initialdir=r'E:\02 源码',
                                            filetypes=[('文本文档', '.txt'),
                                                       ('Excel', '.xls .xlsx'),
                                                       ('All Files', ' *')],
                                            defaultextension='.png')
        print(file)


if __name__ == '__main__':
    test = FileDiaLogDemo()
    test.get_save_file_name()

测试结果(参考):

C:/Users/Administrator/Desktop/Temp/1.txt

在这里插入图片描述

1.6 获取保存文件的属性:asksaveasfile()

import tkinter
from tkinter import filedialog


class FileDiaLogDemo(object):
    def __init__(self):
        window = tkinter.Tk()
        window.withdraw()  # 不显示窗体

    def get_save_file(self):
        file = filedialog.asksaveasfile(title="请选择文件存储路径",
                                        initialdir=r'E:\02 源码',
                                        filetypes=[('文本文档', '.txt'),
                                                   ('Excel', '.xls .xlsx'),
                                                   ('All Files', ' *')],
                                        defaultextension='.png')
        print(f'文件的名称:{file.name}')
        print(f'文件的模式:{file.mode}')
        print(f'文件的编码:{file.encoding}')


if __name__ == '__main__':
    test = FileDiaLogDemo()
    test.get_save_file()

测试结果(参考):

文件的名称:C:/Users/Administrator/Desktop/Temp/3.txt
文件的模式:w
文件的编码:cp936

2 目录

2.1 获取目录名称:askdirectory()

import tkinter
from tkinter import filedialog


class FileDiaLogDemo(object):
    def __init__(self):
        window = tkinter.Tk()
        window.withdraw()  # 不显示窗体

    def get_directory_name(self):
        # 获取单个目录名称(绝对路径)
        directory = filedialog.askdirectory()

        print(directory)


if __name__ == '__main__':
    test = FileDiaLogDemo()
    test.get_directory_name()

测试结果(参考):

C:/Users/Administrator/Desktop/Temp/01

文章来源:https://blog.csdn.net/qq_34745941/article/details/134677624
本文来自互联网用户投稿,该文观点仅代表作者本人,不代表本站立场。本站仅提供信息存储空间服务,不拥有所有权,不承担相关法律责任。