Discussion :

[tamtam64]

Bonjour,

Je souhaiterais une explication si possible sur un code que je ne maitrise pas et que je voudrais comprendre :

Code :

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Public Nom As String
 
 
Declare Function OpenProcess Lib "kernel32" _
                             (ByVal dwDesiredAccess As Long, _
                              ByVal bInheritHandle As Long, _
                              ByVal dwProcessId As Long) As Long
 
Declare Function GetExitCodeProcess Lib "kernel32" _
                                    (ByVal hProcess As Long, _
                                     lpExitCode As Long) As Long
 
Public Const PROCESS_QUERY_INFORMATION = &H400
Public Const STILL_ACTIVE = &H103
 
Public Sub ShellAndWait(ByVal PathName As String, Optional WindowState)
 
    Dim hProg As Long
    Dim hProcess As Long, ExitCode As Long
 
    If IsMissing(WindowState) Then WindowState = 1
    hProg = Shell(PathName, WindowState)
 
    hProcess = OpenProcess(PROCESS_QUERY_INFORMATION, False, hProg)
    Do
 
        GetExitCodeProcess hProcess, ExitCode
        DoEvents
 
    Loop While ExitCode = STILL_ACTIVE
End Sub
 
Sub Dezipper()
 
    Dim CheminZip As String
    Dim DossierUnZip As String
    Dim FichierZip As Variant, ShellStr As String
 
    CheminZip = "C:\program files\7-Zip\"
    If Right(CheminZip, 1) <> "\" Then
        CheminZip = CheminZip & "\"
    End If
 
    If Dir(CheminZip & "7z.exe") = "" Then
        Exit Sub
    End If
 
 
    DossierUnZip = "C:\Divers\Archives\"
    FichierZip = "C:\Divers\Archives\" & Nom
    ShellStr = CheminZip & "7z.exe x -aoa -r" _
             & " " & Chr(34) & FichierZip & Chr(34) _
             & " -o" & Chr(34) & DossierUnZip & Chr(34) & " " & "*.*"
 
    ShellAndWait ShellStr, vbHide
 
End Sub

Comme je ne le maitrise pas je n'arrive pas a degager le nom complet du fichier dézipper, et j'ai besoin de le connaitre car il peut etre differnt du nom du zip .

Merci

[tamtam64]

Je me permet de reposer la meme question dans ce code concernant la partie soulignée: c'est sensiblement identique , mais le e-aot ne signifie rien pour moi

Code :

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Option Explicit
 
Private Declare Function GetExitCodeProcess Lib "kernel32" _
    (ByVal hProcess As Long, lpExitCode As Long) As Long
 
Private Declare Function OpenProcess Lib "kernel32" _
    (ByVal dwDesiredAccess As Long, ByVal bInheritHandle As Long, ByVal dwProcessId As Long) _
    As Long
 
Private Declare Function CloseHandle Lib "kernel32" _
    (ByVal hObject As Long) As Long
 
Private Const PROCESS_QUERY_INFORMATION = &H400
Private Const STILL_ACTIVE = &H103
 
Public UZipDownload As Boolean
 
Private Sub ShellAndWait(ByVal PathName As String, Optional WindowState)
 
Dim hProg As Long
Dim hProcess As Long, ExitCode As Long
 
    UZipDownload = True
 
    If IsMissing(WindowState) Then WindowState = 1
 
    'hProg est un "process ID under Win32"
    hProg = Shell(PathName, WindowState)
 
    'To get the process handle:
    hProcess = OpenProcess(PROCESS_QUERY_INFORMATION, False, hProg)
 
    Do
        'populate Exitcode variable
        GetExitCodeProcess hProcess, ExitCode
        DoEvents
    Loop While ExitCode = STILL_ACTIVE
 
 
    If ExitCode <> 0 Then 
        UZipDownload = False 
        Select Case ExitCode
            Case 1
                MsgBox "Erreur 7z.exe code " & ExitCode & ": Warning (Non fatal error(s))"
            Case 2
                MsgBox "Erreur 7z.exe code " & ExitCode & ": Fatal Error"
            Case 7
                MsgBox "Erreur 7z.exe code " & ExitCode & ": Command line error"
            Case 8
                MsgBox "Erreur 7z.exe code " & ExitCode & ": Not enough memory for operation"
            Case 255
                MsgBox "Erreur 7z.exe code " & ExitCode & ": User stopped the process"
        End Select
    End If
 
 
    hProg = CloseHandle(hProcess)
End Sub
 
'With this example you unzip a fixed zip file: ZipLongName = "C:\Users\Ron\Test.zip"
'Note this file must exist, this is the only thing that you must change before you test it
'The zip file will be unzipped in a new folder in: Application.DefaultFilePath
'Normal if you have not change it this will be your Documents folder
'The name of the folder that the code create in this folder is the Date/Time
'You can change this folder to this if you want to use a fixed folder:
'DestinationPath = "C:\Users\Ron\TestFolder\"
'Read the comments in the code about the commands/Switches in the ShellStr
 
Sub UnzipFile_7z(ZipLongName As String, DestinationPath As String, ProgramName As String, ByVal ProgramPath As String)
 
Dim ShellStr As String
 
    If Right(ProgramPath, 1) <> "\" Then ProgramPath = ProgramPath & "\"
 
    If Dir(ProgramPath & ProgramName) = "" Then
        MsgBox ProgramName & " not found in " & ProgramPath
        Exit Sub
    End If
 
    'extrait (e) toute (*.*)l'archive, renomme l'existant(-aot)
    ShellStr = ProgramPath & ProgramName & " e -aot" & " " & Chr(34) & ZipLongName & _
        Chr(34) & " -o" & Chr(34) & DestinationPath & Chr(34) & " " & "*.*"
 
    Call ShellAndWait(ShellStr, 1)
 
    If UZipDownload = False Then
        MsgBox "Unzip Failure"
    End If
 
End Sub

[kiki29]

Salut, regarde ici

-aot auto rename existing file (for example, name.txt will be renamed to name_1.txt).

[Invité]

Bonjour,
Je ne comprends pas l'intérêt de connaitre le nom du fichier avant de le deziper!
Si tu créé un répertoire temporaire dans lequel tu dezip ton fichier un simple dir te donnera son nom
rep=environ("userprofil") & "\temp" & Format(Now,"yyyy-mm-dd-hh-mm-ss")

[tamtam64]

Bonjour,

Merci pour vos reponses, ce n'est pas si évident, je veux dire quand on ne gere pas forcement les disks comme moi , mais si tu as un tuto sur ca je suis preneur, pour mieux comprendre.

Moipersonnelement je prefererais avoir la main sur le nom du fichier qui est contenu, plutot que le chercher en realite