根据进程号获取进程路径函数（原创）：

这个问题在论坛上被很多人问过，我这里写个函数，公开源代码：

对应CSDN论坛帖子为：

http://community.csdn.net/Expert/topic/4035/4035607.xml?temp=.353039

源代码：

Private Declare Function OpenProcess Lib "kernel32.dll" (ByVal dwDesiredAccess As Long, ByVal bInheritHandle As Long, ByVal dwProcessId As Long) As Long
Private Declare Function EnumProcessModules Lib "psapi.dll" (ByVal hProcess As Long, ByRef lphModule As Long, ByVal cb As Long, ByRef cbNeeded As Long) As Long
Private Declare Function GetModuleFileNameExA Lib "psapi.dll" (ByVal hProcess As Long, ByVal hModule As Long, ByVal ModuleName As String, ByVal nSize As Long) As Long
Private Declare Function CloseHandle Lib "kernel32.dll" (ByVal hObject As Long) As Long

'根据进程号获取进程路径函数，原创：
Function GetProcessPathByProcessID(PID As Long) As String
    On Error GoTo Z
    Dim cbNeeded As Long
    Dim szBuf(1 To 250) As Long
    Dim Ret As Long
    Dim szPathName As String
    Dim nSize As Long
    Dim hProcess As Long
    hProcess = OpenProcess(&H400 Or &H10, 0, PID)
    If hProcess <> 0 Then
        Ret = EnumProcessModules(hProcess, szBuf(1), 250, cbNeeded)
        If Ret <> 0 Then
            szPathName = Space(260)
            nSize = 500
            Ret = GetModuleFileNameExA(hProcess, szBuf(1), szPathName, nSize)
            GetProcessPathByProcessID = Left(szPathName, Ret)
        End If
    End If
    Ret = CloseHandle(hProcess)
    If GetProcessPathByProcessID = "" Then
       GetProcessPathByProcessID = "SYSTEM"
    End If
    Exit Function
Z:
End Function

'我的系统VB进程ID为1716
'下面的结果显示了VB进程的路径为：D:/Microsoft Visual Studio/VB98/VB6.EXE

Private Sub Command1_Click()
    MsgBox GetProcessPathByProcessID(1716)
End Sub

'-------------------------------------------
' 转载请注明出处
' 作者：唐细刚
' 邮箱：tanaya@163.com
'-------------------------------------------