检查如果一个应用程序是空闲的时间段并锁定

Question

在我的项目，我需要一个应用程序锁（像windows一样锁定）。如果应用程序是空闲一段时间的应用程序应锁定即，将显示该应用程序的登录窗口。我如何在WPF C＃应用程序中执行此操作？

Answer 1

您可以使用这些功能

• LockWorkStation
• GetLastInputInfo

看到这个代码，您必须将计时器添加到您的形式，并设置this.timer1.Enabled = TRUE;

using System; 
using System.Collections.Generic; 
using System.ComponentModel; 
using System.Data; 
using System.Drawing; 
using System.Text; 
using System.Windows.Forms; 
using System.Runtime.InteropServices; 

namespace WindowsFormsApplication9 
{ 
    internal struct LASTINPUTINFO 
    { 
    public uint cbSize;  
    public uint dwTime; 
    } 

    public partial class Form1 : Form 
    { 

    [DllImport("User32.dll")] 
    public static extern bool LockWorkStation(); 
    [DllImport("User32.dll")] 
    private static extern bool GetLastInputInfo(ref LASTINPUTINFO Dummy); 
    [DllImport("Kernel32.dll")] 
    private static extern uint GetLastError(); 

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); 
} 

public static long GetTickCount() 
{ 
    return Environment.TickCount; 
} 

public static long GetLastInputTime() 
{ 
    LASTINPUTINFO LastUserAction = new LASTINPUTINFO(); 
    LastUserAction.cbSize = (uint)System.Runtime.InteropServices.Marshal.SizeOf(LastUserAction); 
    if (!GetLastInputInfo(ref LastUserAction)) 
    { 
    throw new Exception(GetLastError().ToString()); 
    } 

    return LastUserAction.dwTime; 
} 

    public Form1() 
    { 
     InitializeComponent(); 
    } 

    private void timer1_Tick(object sender, EventArgs e) 
    { 
     if (GetIdleTime() > 10000) //10 secs, Time to wait before locking 
     LockWorkStation(); 
    } 

    private void Form1_Load(object sender, EventArgs e) 
    { 
     timer1.Start(); 
    } 
    } 
} 

Answer 2

设置每一个“主动”动作发生的时间（你需要连接到他们）负载超时，并且，复位定时器回到开始。

Answer 3

IMO接受的答案是不是因为这个方法好：

http://www.codeproject.com/Articles/30345/Application-Idle

CodeProject上的文章使用Windows消息，这将导致组件考虑该申请没有闲着，如

public enum ActivityMessages : int 
{ 
    /// <summary> 
    /// Cursor moved while within the nonclient area. 
    /// </summary> 
    WM_NCMOUSEMOVE = 0x00A0, 
    /// <summary> 
    /// Mouse left button pressed while the cursor was within the nonclient area. 
    /// </summary> 
    WM_NCLBUTTONDOWN = 0x00A1, 
    /// <summary> 
    /// Mouse left button released while the cursor was within the nonclient area. 
    /// </summary> 
    WM_NCLBUTTONUP = 0x00A2, 
    /// <summary>