如何在 VB6 中读取子进程的标准输出？

Question

在 VB6 中创建进程时（与 这个问题:)，我使用以下结构：

Private Type STARTUPINFO
      cb As Long
      lpReserved As String
      lpDesktop As String
      lpTitle As String
      dwX As Long
      dwY As Long
      dwXSize As Long
      dwYSize As Long
      dwXCountChars As Long
      dwYCountChars As Long
      dwFillAttribute As Long
      dwFlags As Long
      wShowWindow As Integer
      cbReserved2 As Integer
      lpReserved2 As Long
      hStdInput As Long
      hStdOutput As Long
      hStdError As Long
   End Type

在开始我的流程之前，需要对 STARTUPINFO.hStdOutput 进行哪些操作才能让我的 VB6 应用程序读取托管流程的输出？

谢谢！！

Answer 1

跟进这个其他问题OP，我发布了一种替代方法来执行命令并获取标准输出：

' References: "Windows Script Host Shell Object Model" '

Public Declare Sub Sleep Lib "kernel32" Alias "Sleep" ( _
  ByVal dwMilliseconds As Long)

Function ExecuteCommand(cmd As String, ExpectedResult as Long) As String
  Dim shell As New IWshRuntimeLibrary.WshShell
  Dim exec As IWshRuntimeLibrary.WshExec

  Set exec = shell.Exec(cmd)
  While exec.Status = 0
     Sleep 100
  Wend

  If exec.ExitCode = ExpectedResult Then
    ExecuteCommand = exec.StdOut.ReadAll
  Else
    ExecuteCommand = vbNullString     ' or whatever '
  End
End Function

Answer 2

Microsoft 在此处提供了有关如何执行此操作的示例。

Answer 3

请参阅AttachConsole(ATTACH_PARENT_PROCESS)