Question

试图创造一个魔兽世界＆＃34;发射器。

它应该如何运作：按＆＃34;启动Authserver＆＃34;你可以找到authserver.exe。然后保存位置，这样你就不必两次这样做了。一旦按下，显然应该启动authserver.exe。

会发生什么：按＆＃34;启动Authserver＆＃34;它会打开并立即关闭并显示错误：错误，无法打开导致authserver.conf无法找到。

注意： Authserver.exe和authserver.conf位于同一个文件夹中，我可以手动启动它。

Mort昨天帮助我解决了类似的问题，我希望相同的VB代码可以工作，因为这看起来很明显。

Private filePath As String = String.Empty

Private Sub PlayButton_Click(sender As System.Object, e As System.EventArgs) Handles PlayButton.Click
    Try
        If filePath.Length = 0 Then
            Dim diagResult As DialogResult = OpenFileDialog1.ShowDialog()
            If diagResult = Windows.Forms.DialogResult.OK Then
                filePath = OpenFileDialog1.FileName
                If filePath.ToUpper.EndsWith("WOW.EXE") Then
                    Process.Start(filePath)
                Else
                    MessageBox.Show("Wrong file selected!")
                    filePath = String.Empty
                End If
            End If
        Else
            Process.Start(filePath)
        End If

    Catch ex As Exception
        MessageBox.Show(String.Concat("An error occurred in the play button click:", ex.Message))
    End Try

End Sub

有人有个主意吗？提前：谢谢。

Answer 1

尝试这样的事情：

＆＃13;

    Private Sub PlayButton_Click(sender As System.Object, e As System.EventArgs) Handles PlayButton.Click
        Try
            OpenFileDialog1.FileName = ""
            OpenFileDialog1.Filter = "World of Warcraft (WOW.EXE)|WOW.EXE"
            Dim diagResult As DialogResult = OpenFileDialog1.ShowDialog
            If diagResult = Windows.Forms.DialogResult.OK Then
                Dim p As New Process
                Dim fn As New System.IO.FileInfo(OpenFileDialog1.FileName)
                p.StartInfo.WorkingDirectory = fn.DirectoryName
                p.StartInfo.FileName = fn.Name
                p.Start()
            End If
        Catch ex As Exception
            MessageBox.Show(String.Concat("An error occurred in the play button click:", ex.Message))
        End Try
    End Sub

＆＃13;

无法启动.exe

1 个答案: