我正在尝试使用新的进程ID启动我的应用程序的单独实例。有没有办法做到这一点?以下代码将显示单独的表单,但它共享原始表单的进程ID:
Private Sub NewToolStripMenuItem_Click(sender As System.Object, e As System.EventArgs) Handles NewToolStripMenuItem.Click
Dim SecondForm As New MyForm()
SecondForm.Show()
End Sub
当用户从工具条菜单中选择“新建”选项时,我基本上想要复制与打开应用程序相同的行为。
答案 0 :(得分:0)
我找到了解决方案。
Private Sub NewToolStripMenuItem_Click(sender As System.Object, e As System.EventArgs) Handles NewToolStripMenuItem.Click
Dim ExePath As String = Application.ExecutablePath
Process.Start(ExePath)
End Sub