使用python自动从windows文件对话框打开文件

Open file from windows file dialog with python automatically(使用python自动从windows文件对话框打开文件)

问题描述

我进行自动化测试并获得一个文件对话框.我想用 python 或 selenium 从 windows 打开文件对话框中选择一个文件.

注意:该对话框由其他程序提供.我不想用 Tkinter 创建它.

窗口看起来像:

.

如何做到这一点?

考虑使用

代码示例，在记事本中打开文件.请注意，语法取决于区域设置(它使用 GUI 程序中可见的窗口标题/控件标签):

来自 pywinauto 导入应用程序app = application.Application().start_('notepad.exe')app.Notepad.MenuSelect('文件->打开')# app.[窗口标题].[控件名称]...app.Open.Edit.SetText('filename.txt')app.Open.Open.Click()

I do automated testing and get a file dialog. I want to choose a file from the windows open file dialog with python or selenium.

NOTE: The dialog is given by an other program. I don't want to create it with Tkinter.

The Window looks like:

.

How to do this?

Consider using the pywinauto package. It has a very natural syntax to automate any GUI programs.

Code example, opening a file in notepad. Note that the syntax is locale dependent (it uses the visible window titles / control labels in your GUI program):

from pywinauto import application
app = application.Application().start_('notepad.exe')
app.Notepad.MenuSelect('File->Open')
# app.[window title].[control name]...
app.Open.Edit.SetText('filename.txt')
app.Open.Open.Click()

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