最简单的方法
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If UBound(Diagnostics.Process.GetProcessesByName(Diagnostics.Process.GetCurrentProcess.ProcessName)) 0 Then End
前提是你的程序名别太普通了容易和别的进程重名就不行了
方法一：通过Diagnostics.Process.GetProcessesByName函数来检测程序是否已经启动
Imports System.Windows.Forms
Module Module1
Sub Main()
'检测多重启动
If Diagnostics.Process.GetProcessesByName( _
Diagnostics.Process.GetCurrentProcess.ProcessName).Length 1 Then
MessageBox.Show("已经一个实例的本程序正在运行。")
Return
End If
Application.Run(New Form())
End Sub
End Module
不过这个方法有个缺点，如果用户改了下exe的名字就检测不到了，所以更好的方法如下
方法2：使用Mutex
Imports System.Windows.Forms
Module Module1
Sub Main()
Dim createdNew As Boolean
' 创建mutex
Dim mutex As System.Threading.Mutex = _
New System.Threading.Mutex(True, "YourAppName", createdNew)
If createdNew = False Then
MessageBox.Show("已经一个实例的本程序正在运行。")
Return
End If
Application.Run(New Form())
' 释放mutex
mutex.ReleaseMutex()
End Sub
End Module
试试看这样行不：在应用程序设置中，勾选“生成单个实例应用程序”，然后在应用程序事件中处理这个事件
Private Sub MyApplication_StartupNextInstance(sender As Object, e As StartupNextInstanceEventArgs) Handles Me.StartupNextInstance
'这里的 e.CommandLine应该就是双击第二个文件时传进来的命令行
End Sub
在项目上右键，然后选择属性，打开如下图所示的项目属性对话框：
然后选择左边的“应用程序”标签页，再勾选上“生成单个实例应用程序”。
Public Class Form1
Dim Files() As String
Dim ImgNum As Integer = 0
Dim CurIdx As Integer = 0
Sub OpenFiles()
InitialSet()
Try
If OpenFileDialog1.ShowDialog = Windows.Forms.DialogResult.OK Then
If Not OpenFileDialog1.FileNames Is Nothing Then
Files = OpenFileDialog1.FileNames
Else
Files(0) = OpenFileDialog1.FileName
End If
ImgNum = Files.Length
Dim s As String = String.Format("你共打开 {0} 张图片", ImgNum)
Me.Text = s
StateSet()
CurIdx = 1
DisplayImage(CurIdx)
Else
StateSet()
Me.PictureBox1.Image = Nothing
End If
Catch ex As Exception
MsgBox(ex.Message)
End Try
End Sub
Sub InitialSet()
Me.OpenFileDialog1.Multiselect = True
Me.OpenFileDialog1.Filter = "(*.jpg)|*jpg"
Me.OpenFileDialog1.FilterIndex = 0
ImgNum = 0
CurIdx = 0
If Not Files Is Nothing Then
Array.Clear(Files, 0, Files.Length - 1)
End If
End Sub
Sub DisplayImage(ByRef Idx As Integer)
If IO.Path.GetExtension(Files(Idx - 1)).ToLower = ".jpg" Then
Me.PictureBox1.Load(Files(Idx - 1))
End If
StateSet()
End Sub
Sub StateSet()
If ImgNum 1 Then
Me.btnNext.Enabled = True
Else
Me.btnNext.Enabled = False
Me.btnPervious.Enabled = False
Exit Sub
End If
If CurIdx 1 And ImgNum 1 Then
Me.btnPervious.Enabled = True
Else
Me.btnPervious.Enabled = False
End If
If CurIdx ImgNum And ImgNum 1 Then
Me.btnNext.Enabled = True
Else
Me.btnNext.Enabled = False
End If
End Sub
Sub NextIamge(ByRef Idx As Integer)
If Idx ImgNum Then
Idx += 1
Else
Exit Sub
End If
DisplayImage(Idx)
End Sub
Sub PerviousIagme(ByRef Idx As Integer)
If Idx 1 Then
Idx -= 1
Else
Exit Sub
End If
DisplayImage(Idx)
End Sub
Private Sub btnNext_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles btnNext.Click
Try
NextIamge(CurIdx)
Catch ex As Exception
MsgBox(ex.Message)
End Try
End Sub
Private Sub btnOpen_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles btnOpen.Click
OpenFiles()
End Sub
Private Sub btnPervious_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles btnPervious.Click
Try
PerviousIagme(CurIdx)
Catch ex As Exception
MsgBox(ex.Message)
End Try
End Sub
End Class
vb.net没用过，vb6.0的话可以用getobject这种方法，你可以改改试试，看能不能用
Dim xlApp As Object
Set xlApp = GetObject(ExcelFileName).Application '获取Excel对象
