Question

我希望Visual Basic能够在目录“C：\ projectTest \”上运行“make”命令。

我试着用这个：

    Dim output As String = String.Empty

    Using Process As New Process
        Process.StartInfo = New ProcessStartInfo("cmd")
        Process.StartInfo.WorkingDirectory = "C:\projectTest\"
        Process.StartInfo.UseShellExecute = False
        Process.StartInfo.CreateNoWindow = True
        Process.StartInfo.RedirectStandardInput = True
        Process.StartInfo.RedirectStandardOutput = True
        Process.StartInfo.RedirectStandardError = True
        Process.Start()
        Process.BeginOutputReadLine()
        AddHandler Process.OutputDataReceived,
 _
           Sub(processSender As Object, lineOut As DataReceivedEventArgs)
               output += lineOut.Data + vbCrLf
           End Sub

        Using InputStream As System.IO.StreamWriter = Process.StandardInput
            InputStream.AutoFlush = False
            InputStream.WriteLine("make")
        End Using
        Do
            Application.DoEvents()
        Loop Until Process.HasExited
    End Using

此代码能够捕获控制台的“gcc ...”部分（来自Makefile），但不会捕获错误（如果我手动打开cmd并在该目录上运行make，则会弹出错误）

如何捕获显示的所有内容，包括错误？

Answer 1

不止一个问题。首先，正如@ shf301已经告诉过你的那样，你忘了阅读stderr。他反过来忘了添加额外的一行：

    Process.Start()
    AddHandler Process.OutputDataReceived, _
       Sub(processSender As Object, lineOut As DataReceivedEventArgs)
           output += lineOut.Data + vbCrLf
       End Sub
    Process.BeginOutputReadLine()
    AddHandler Process.ErrorDataReceived, _
       Sub(processSender As Object, lineOut As DataReceivedEventArgs)
           output += lineOut.Data + vbCrLf
       End Sub
    Process.BeginErrorReadLine()

还有一个非常麻烦的问题，你的事件处理程序运行得很晚。在进程已经退出之后，它们会激活。这些处理程序在线程池线程上运行的副作用。在使用输出变量之前，您需要等待任意（且不可饶恕）的时间：

Do Application.DoEvents() Loop Until Process.HasExited System.Threading.Thread.Sleep(1000)

这太难看了。按照任何 IDE或编辑器的方式执行此操作。将输出重定向到临时文件，然后读取文件：

Dim tempfile As String = System.IO.Path.GetTempFileName Using Process As New Process Process.StartInfo = New ProcessStartInfo("cmd.exe") Process.StartInfo.Arguments = "/c make 1> """ + tempfile + """ 2>&1" Process.StartInfo.WorkingDirectory = "C:\projectTest" Process.StartInfo.UseShellExecute = False Process.StartInfo.CreateNoWindow = True Process.Start() Process.WaitForExit() output = System.IO.File.ReadAllText(tempfile) System.IO.File.Delete(tempfile) End Using

使用mystic命令行进行一些注释：

/ c告诉cmd.exe只执行单个命令，然后退出

1＆GT;将输出重定向到临时文件

2＆gt;＆amp; 1告诉cmd.exe将stderr重定向到stdout

三重双引号确保临时文件名中的空格不会造成麻烦。

同样的2>&1也会修复原来的问题;）

Answer 2

错误通常写入StandardError流，您只读取StandardOutput流。为ErrorDataReceived事件添加事件处理程序，您应该会看到错误。

AddHandler Process.ErrorDataReceived, _
       Sub(processSender As Object, lineOut As DataReceivedEventArgs)
           output += lineOut.Data + vbCrLf
       End Sub

Answer 3

对于像我这样的人，我们想知道为什么我们无法在异步模式下获得错误。添加行

process.BeginErrorReadLine()

Visual Basic捕获cmd的输出

3 个答案: