如果没有移动，则在 5 秒后(空闲)关闭 winform 应用程序中的会话 [英] close session in winform app after 5 seconds(idle) if no movement

问题描述

我正在编写一个 c# Winform 应用程序，如果计算机空闲 5 秒，我需要关闭会话.该应用程序就像一个餐厅应用程序，当服务员打开他的会话时，我会在 5 秒后关闭它.

I am writing a c# Winform application and I need to close session if the computer is idle for 5 seconds. The application is like a restaurant application, when the waiter leaves his session open, I will close it after 5 seconds.

我找到了一些代码，但我不知道如何使用它以及如何触发它

I found some code but I dont know how to use it and how to trigger it

using System.Runtime.InteropServices;

[DllImport("User32.dll")]
private static extern bool GetLastInputInfo(ref LASTINPUTINFO plii);   

internal struct LASTINPUTINFO
{
    public uint cbSize;

    public uint dwTime;
}

有人可以帮我吗?

推荐答案

请按照以下步骤操作:

1- 将 Timer 添加到您的 Form.

1- Add a Timer to your Form.

2- 将其间隔属性设置为 1000(在 form_load 或在属性窗口的设计模式中设置).

2- Set its interval property to 1000 (set it in form_load or in design mode from Properties window).

3-将此方法添加到您的 Form 类中.

3-Add this method to your Form class.

public static uint GetIdleTime()
{
     LASTINPUTINFO LastUserAction = new LASTINPUTINFO();
     LastUserAction.cbSize = (uint)System.Runtime.InteropServices.Marshal.SizeOf(LastUserAction);
     GetLastInputInfo(ref LastUserAction);
     return ((uint)Environment.TickCount - LastUserAction.dwTime);
}

4-在Form_Load中启动定时器:

timer1.Start();

5- 在定时器 tick 事件检查 GetIdleTime()，例如如果它大于 5000 表示应用程序空闲 5 秒之前.

5- in timer tick event check GetIdleTime(), for example if it is greater than 5000 means application was idled since 5 seconds ago.

private void timer1_Tick(object sender, EventArgs e)
{
    if (GetIdleTime() > 5000)  
       Application.Exit();//For Example
}

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