Question

有没有办法注册在特定文件名的可执行文件启动时触发的事件？我知道通过获取进程句柄并注册退出事件，在进程退出时获取事件很容易。但是，当一个尚未运行的进程启动时，如何在不轮询所有正在运行的进程的情况下通知您如何通知？

Answer 1

您可以使用以下内容：

    private ManagementEventWatcher WatchForProcessStart(string processName)
    {
        string queryString =
            "SELECT TargetInstance" +
            "  FROM __InstanceCreationEvent " +
            "WITHIN  10 " +
            " WHERE TargetInstance ISA 'Win32_Process' " +
            "   AND TargetInstance.Name = '" + processName + "'";

        // The dot in the scope means use the current machine
        string scope = @"\\.\root\CIMV2";

        // Create a watcher and listen for events
        ManagementEventWatcher watcher = new ManagementEventWatcher(scope, queryString);
        watcher.EventArrived += ProcessStarted;
        watcher.Start();
        return watcher;
    }

    private ManagementEventWatcher WatchForProcessEnd(string processName)
    {
        string queryString =
            "SELECT TargetInstance" +
            "  FROM __InstanceDeletionEvent " +
            "WITHIN  10 " +
            " WHERE TargetInstance ISA 'Win32_Process' " +
            "   AND TargetInstance.Name = '" + processName + "'";

        // The dot in the scope means use the current machine
        string scope = @"\\.\root\CIMV2";

        // Create a watcher and listen for events
        ManagementEventWatcher watcher = new ManagementEventWatcher(scope, queryString);
        watcher.EventArrived += ProcessEnded;
        watcher.Start();
        return watcher;
    }

    private void ProcessEnded(object sender, EventArrivedEventArgs e)
    {
        ManagementBaseObject targetInstance = (ManagementBaseObject) e.NewEvent.Properties["TargetInstance"].Value;
        string processName = targetInstance.Properties["Name"].Value.ToString();
        Console.WriteLine(String.Format("{0} process ended", processName));
    }

    private void ProcessStarted(object sender, EventArrivedEventArgs e)
    {
        ManagementBaseObject targetInstance = (ManagementBaseObject)e.NewEvent.Properties["TargetInstance"].Value;
        string processName = targetInstance.Properties["Name"].Value.ToString();
        Console.WriteLine(String.Format("{0} process started", processName));
    }

然后，您将调用WatchForProcessStart和/或WatchForProcessEnd传入您的进程名称（例如“notepad.exe”）。

ManagementEventWatcher对象是从两个Watch *方法返回的，因为它实现了IDisposable，所以你应该在完成这些对象时调用Dispose来防止出现问题。

如果您需要在流程启动后更快地引发事件，您还可以更改查询中的轮询值。要执行此操作，请将“WITHIN 10”行更改为少于10的内容。

Answer 2

WMI可以在创建进程时创建事件。然后，您可以过滤这些事件。

Answer 3

这是代码。

请注意，您必须像管理员一样启动Visual Studio才能执行此代码。

using System;
using System.Management;

namespace AppLaunchDetector
{
    class Program
    {
        static void Main(string[] args)
        {           
            ManagementEventWatcher w = null;
            WqlEventQuery q;
            try
            {
                q = new WqlEventQuery();
                q.EventClassName = "Win32_ProcessStartTrace";
                w = new ManagementEventWatcher(q);
                w.EventArrived += new EventArrivedEventHandler(ProcessStartEventArrived);
                w.Start();
                Console.ReadLine(); // block main thread for test purposes
            }
            catch (Exception ex)
            {

            }
            finally
            {
                w.Stop();
            }
        }

        static void ProcessStartEventArrived(object sender, EventArrivedEventArgs e)
        {
            foreach (PropertyData pd in e.NewEvent.Properties)
            {
                Console.WriteLine("\n============================= =========");
                Console.WriteLine("{0},{1},{2}", pd.Name, pd.Type, pd.Value);
            }
        }
    }
}

用于Process可执行文件的.NET事件

3 个答案: