VB.NET2010窗口服务运行实际应用程序，而不仅仅是进程

我正在尝试创建一个服务来检查Microsoft Lync是否正在运行。如果它正在运行，则只需写入事件日志。如果未运行，请运行exe，然后登录到Lync。 我遇到的问题是，当需要运行exe时，它会去启动进程，但它从来没有实际运行过应用程序。我试图看看是否可以使用记事本，但它所做的只是在任务管理器中创建流程，但从未打开实际的应用程序

  Imports System
  Imports System.Data
  Imports System.Timers
  Imports System.Diagnostics
  Imports System.Data.SqlClient
  Imports System.ServiceProcess
  Imports System.Windows.Forms

    Public Class Service1

    Protected Overrides Sub OnStart(ByVal args() As String)
        EventLog.WriteEntry("In Onstart", "starting timer")
           Timer1.Start()
    End Sub



   Protected Overrides Sub OnStop()
   End Sub

   Private Sub Timer1_Elapsed(ByVal sender As System.Object, 
   ByVal e As System.Timers.ElapsedEventArgs) Handles Timer1.Elapsed

       If IsProcessRunning("communicator") Then
       EventLog.WriteEntry("no problem")
       Else
       EventLog.WriteEntry("not running")
       Dim info As New ProcessStartInfo("C:\Program Files (x86)\Microsoft Lync\communicator.exe")
       info.UseShellExecute = False
       info.RedirectStandardError = True
       info.RedirectStandardInput = True
       info.RedirectStandardOutput = True
       info.CreateNoWindow = True
       info.ErrorDialog = False
       info.WindowStyle = ProcessWindowStyle.Hidden

     Dim process__1 As Process = Process.Start(info)

     End If

   End Sub

   Public Function IsProcessRunning(ByVal name As String) As Boolean
      For Each clsProcess As Process In Process.GetProcesses()
        If clsProcess.ProcessName.StartsWith(name) Then
         Return True
      End If
      Next
         Return False
   End Function
End Class

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问题来自这样一个事实：在Windows中（特别是6.x版之后），服务运行在完全隔离的会话和桌面上，没有任何用户交互的机会，出于安全原因，这是出于设计，不鼓励这样做。你正在启动的程序确实会启动，但它是在隐藏的桌面上启动的（与记事本相同），在那里用户永远看不到它

快速而棘手的解决方法是在控制面板中将服务标记为interactive，然后启动interactive services检测服务。执行此操作时，当您的服务运行该程序时，任务栏中将闪烁一个窗口，告诉您有来自服务的消息，以便您可以切换到并行桌面并实际看到它。这对用户来说非常不方便，并且被广泛认为是一种不良做法

真正的解决方案是使程序成为常规应用程序而不是服务，并通过Windows中的某个autostart位置为每个用户运行它。它不需要有可见的UI，而是在与您相同的上下文中运行。或者离开服务，但也在用户空间中放置一些应用程序，这些应用程序与服务通信只是为了运行第二个程序。在任何情况下，一般规则是永远不要从服务流程进行任何类型的用户交互

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这里有一篇文章解释了这些问题和一些解决方法

我能够创建一个程序来检测Lync是否正在运行，如果没有，请重新启动程序

Imports System.Diagnostics
Imports Microsoft.Lync
Imports Microsoft.Lync.Model
Imports Microsoft.Lync.Model.Conversation
Imports Microsoft.Lync.Model.Group

Public Class frmCheckLync
Private _lyncClient As LyncClient
Private _contactManager As ContactManager
Private _conversationManager As ConversationManager
Private _self As Self
Private _groups As Dictionary(Of String, Group)
Private _contactSubscriptions As Dictionary(Of String, ContactSubscription)
Public Event StateChanged As EventHandler(Of ClientStateChangedEventArgs)
Public _signIn As IAsyncResult
Public asyncState As Object() = {_lyncClient}
Public clsProcess As Process


Private Sub Form1_Load(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles MyBase.Load
    Try

        Timer1.Enabled = True

        AddHandler Timer1.Elapsed, AddressOf Timer1_Tick

        Timer1.Interval = 600

        Timer1.Start()

    Catch ex As Exception

        Diagnostics.EventLog.WriteEntry("This is a test" + ex.Message.ToString, "This is a test")

    End Try
End Sub

Private Sub Timer1_Tick(ByVal sender As Object, ByVal e As System.EventArgs)

    Try

        Timer1.Stop()

        GetClient()


        Try
            CheckClient()
        Catch ex As Exception

        End Try



    Catch ex As Exception




    Finally

        Timer1.Start()

    End Try

End Sub

Public Function IsProcessRunning(ByVal name As String) As Boolean
    For Each Me.clsProcess In Process.GetProcesses()

        If Me.clsProcess.ProcessName.StartsWith(name) Then
            Return True
        End If
    Next
    Return False
End Function

Public Sub GetClient()
    If IsProcessRunning("communicator") = False Then

        Try
            Dim info As New ProcessStartInfo("C:\Program Files\Microsoft Lync\communicator.exe")
            info.UseShellExecute = False
            info.RedirectStandardError = True
            info.RedirectStandardInput = True
            info.RedirectStandardOutput = True
            info.CreateNoWindow = True
            info.ErrorDialog = False
            info.WindowStyle = ProcessWindowStyle.Normal

            Dim process__1 As Process = Process.Start(info)
        Catch ex1 As Exception

        End Try
        Try
            Dim info1 As New ProcessStartInfo("C:\Program Files (x86)\Microsoft Lync\communicator.exe")
            info1.UseShellExecute = False
            info1.RedirectStandardError = True
            info1.RedirectStandardInput = True
            info1.RedirectStandardOutput = True
            info1.CreateNoWindow = True
            info1.ErrorDialog = False
            info1.WindowStyle = ProcessWindowStyle.Normal

            Dim process__2 As Process = Process.Start(info1)
        Catch ex2 As Exception

        End Try
    End If
End Sub
Public Sub CheckClient()
    Try
        _lyncClient = LyncClient.GetClient()
    Catch
        GetClient()
    End Try


    If _lyncClient.State = ClientState.SignedOut Then
        _signIn = _lyncClient.BeginSignIn(Nothing, Nothing, Nothing,
                                          Function(result)
                                              If result.IsCompleted Then
                                                  _lyncClient.EndSignIn(result)
                                                  ' Setup application logic

                                                  ' could not sign in 
                                              Else
                                              End If

                                          End Function, TryCast("Local user signing in", Object))

    End If

End Sub



End Class