土司风格弹出我的应用程序 [英] toast style popup for my application

问题描述

我创建了在任务栏运行的应用程序。当用户点击它会弹出应用程序等等。我想是类似的功能，在MSN的时候我的一个朋友在记录，显然这是知道作为一个面包弹出？
我基本上想要的东西，从每20分钟土司风格FOM在任务栏的弹出应用程序。

I have created an application that runs in the taskbar. When a user clicks the application it pops up etc. What I would like is similar functionality to that in MSN when one of my friends logs in. Apparently this is know as a toast popup? I basically want something to popup from every 20 minutes toast style fom the application in the taskbar.

我现有的应用程序是基于C＃编写与.NET的WinForms 3.5

My existing application is winforms based written in C# with .net 3.5

干杯

推荐答案

这是非常简单的。你只需要在屏幕外的区域设置窗口动画和它的位置，直到其完全可见。下面是一个示例代码：

This is pretty simple. You just need to set window in off-screen area and animate it's position until it is fully visible. Here is a sample code:

public partial class Form1 : Form
{
    private Timer timer;
    private int startPosX;
    private int startPosY;

    public Form1()
    {
        InitializeComponent();
        // We want our window to be the top most
        TopMost = true;
        // Pop doesn't need to be shown in task bar
        ShowInTaskbar = false;
        // Create and run timer for animation
        timer = new Timer();
        timer.Interval = 50;
        timer.Tick += timer_Tick;
    }

    protected override void OnLoad(EventArgs e)
    {
        // Move window out of screen
        startPosX = Screen.PrimaryScreen.WorkingArea.Width - Width;
        startPosY = Screen.PrimaryScreen.WorkingArea.Height;
        SetDesktopLocation(startPosX, startPosY);
        base.OnLoad(e);
        // Begin animation
        timer.Start();
    }

    void timer_Tick(object sender, EventArgs e)
    {
        //Lift window by 5 pixels
        startPosY -= 5; 
        //If window is fully visible stop the timer
        if (startPosY < Screen.PrimaryScreen.WorkingArea.Height - Height)
            timer.Stop();
        else
           SetDesktopLocation(startPosX, startPosY);
    }
}

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