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python 在其他 windows 顶部打开文件夹

[英]python open folder on top of other windows

I have build in flask app, that recive a path and then open a folder.我已经构建了 flask 应用程序,它接收路径然后打开一个文件夹。 my issue is that the folder is opened minimized and i would like to open it on top of all windows.我的问题是该文件夹已最小化打开,我想在所有 windows 之上打开它。 how can I achieve it?我怎样才能实现它? I'm entering the URL http://localhost:9999/number from other web and it does open the folder just minimize:(我正在输入 URL http://localhost:9999/number 来自其他 web 并打开文件夹只是最小化:(

CLOSE_TAB = "<script>window.onload = window.close();</script>"
@app.route('/<number>')
def open_folder(number):
    try:
        home = str(Path.home())
        full_path = home + '/' + number
        folder_path = os.path.realpath(full_path)
        if os.path.exists(folder_path):
            #Open my folder by number
            os.startfile(folder_path)
            return CLOSE_TAB
        else:
            return FOLDER_NOT_FOUND
    except Exception as error:
        print(error)
        return "{}".format(str(error))


if __name__ == '__main__':
    app.run(host="localhost", port=9999)

Since Python 3.10 the function os.startfile has parameter show_cmd which takes one of the int values documented for Win32's function ShowWindow for parameter nCmdShow .自 Python 3.10 起,function os.startfile具有参数show_cmd ,该参数采用 Win32 的ShowWindow参数nCmdShow记录的 int 值之一

Among others there is其中有

  • 1 for normal 1 为正常
  • 2 for minimized 2 最小化
  • 3 for maximized 3 最大化

It is up to the started application if the argument is respected.如果参数得到尊重,则取决于启动的应用程序。

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