J'essaie de coder quelque chose dans Visual Basic, plus particulièrement Visual Studio 2010. Je veux, en un clic, mon programme exécuter une commande. Est-ce possible?
Oui. Vous pouvez utiliser Process.Start
pour lancer un exécutable, y compris une application console.
Si vous devez lire la sortie de l'application, vous devrez peut-être lire à partir de StandardOutput stream pour pouvoir imprimer quoi que ce soit à partir de l'application que vous lancez.
Voici un exemple:
Process.Start("CMD", "/C Pause")
/C Carries out the command specified by string and then terminates
/K Carries out the command specified by string but remains
Et voici une fonction étendue: (Notez les lignes de commentaires utilisant les commandes CMD.)
#Region " Run Process Function "
' [ Run Process Function ]
'
' // By Elektro H@cker
'
' Examples :
'
' MsgBox(Run_Process("Process.exe"))
' MsgBox(Run_Process("Process.exe", "Arguments"))
' MsgBox(Run_Process("CMD.exe", "/C Dir /B", True))
' MsgBox(Run_Process("CMD.exe", "/C @Echo OFF & For /L %X in (0,1,50000) Do (Echo %X)", False, False))
' MsgBox(Run_Process("CMD.exe", "/C Dir /B /S %SYSTEMDRIVE%\*", , False, 500))
' If Run_Process("CMD.exe", "/C Dir /B", True).Contains("File.txt") Then MsgBox("File found")
Private Function Run_Process(ByVal Process_Name As String, _
Optional Process_Arguments As String = Nothing, _
Optional Read_Output As Boolean = False, _
Optional Process_Hide As Boolean = False, _
Optional Process_TimeOut As Integer = 999999999)
' Returns True if "Read_Output" argument is False and Process was finished OK
' Returns False if ExitCode is not "0"
' Returns Nothing if process can't be found or can't be started
' Returns "ErrorOutput" or "StandardOutput" (In that priority) if Read_Output argument is set to True.
Try
Dim My_Process As New Process()
Dim My_Process_Info As New ProcessStartInfo()
My_Process_Info.FileName = Process_Name ' Process filename
My_Process_Info.Arguments = Process_Arguments ' Process arguments
My_Process_Info.CreateNoWindow = Process_Hide ' Show or hide the process Window
My_Process_Info.UseShellExecute = False ' Don't use system Shell to execute the process
My_Process_Info.RedirectStandardOutput = Read_Output ' Redirect (1) Output
My_Process_Info.RedirectStandardError = Read_Output ' Redirect non (1) Output
My_Process.EnableRaisingEvents = True ' Raise events
My_Process.StartInfo = My_Process_Info
My_Process.Start() ' Run the process NOW
My_Process.WaitForExit(Process_TimeOut) ' Wait X ms to kill the process (Default value is 999999999 ms which is 277 Hours)
Dim ERRORLEVEL = My_Process.ExitCode ' Stores the ExitCode of the process
If Not ERRORLEVEL = 0 Then Return False ' Returns the Exitcode if is not 0
If Read_Output = True Then
Dim Process_ErrorOutput As String = My_Process.StandardOutput.ReadToEnd() ' Stores the Error Output (If any)
Dim Process_StandardOutput As String = My_Process.StandardOutput.ReadToEnd() ' Stores the Standard Output (If any)
' Return output by priority
If Process_ErrorOutput IsNot Nothing Then Return Process_ErrorOutput ' Returns the ErrorOutput (if any)
If Process_StandardOutput IsNot Nothing Then Return Process_StandardOutput ' Returns the StandardOutput (if any)
End If
Catch ex As Exception
'MsgBox(ex.Message)
Return Nothing ' Returns nothing if the process can't be found or started.
End Try
Return True ' Returns True if Read_Output argument is set to False and the process finished without errors.
End Function
#End Region
Ou, vous pouvez le faire de manière très simple.
Dim OpenCMD
OpenCMD = CreateObject("wscript.Shell")
OpenCMD.run("Command Goes Here")
Vous devez utiliser CreateProcess [ http://msdn.Microsoft.com/en-us/library/windows/desktop/ms682425(v=vs.85).aspx ]
Pour ex:
LPTSTR szCmdline [] = _tcsdup (TEXT ("\" C:\Program Files\MonApp\"-L -S")); CreateProcess (NULL, szCmdline, / ... /);