我正在尝试创建一个 zip 文件,然后将三个文件夹复制到其中。我在第 33 行 char 1 上收到错误,需要错误状态对象,我已经搜索和谷歌搜索,但似乎无法理解我正在阅读的内容或理解我真正需要搜索的内容。无论如何,这是我的代码。
Option Explicit
Dim objFSO, objFolder1, objFolder2, objFolder3, FolderToZip, ziptoFile, FolderGroup
Dim ShellApp, eFile, oNewZip, strZipHeader
Dim ZipName, Folder, i, Zip, Item
Set objFSO = CreateObject("Scripting.FileSystemObject")
Set objFolder1 = objFSO.GetFolder("C:\Windows\Temp\SMSTSLog")
Set objFolder2 = objFSO.GetFolder ("C:\Windows\System32\CCM\Logs")
Set objFolder3 = objFSO.GetFolder ("C:\Windows\SysWOW64\CCM\Logs")
'For Each efile In objFolder.Files
' If DateDiff("d",eFile.DateLastModified,Now) >= 2 Then
' objFSO.MoveFile eFile, "C:\Documents and Settings\User\Desktop\Test2\"
' End If
'Next
Wscript.Sleep 2000
Set oNewZip = objFSO.OpenTextFile("C:\win7tools\testing script.zip", 8, True)
strZipHeader = "PK" & Chr(5) & Chr(6)
For i = 0 To 17
strZipHeader = strZipHeader & Chr(0)
Next
oNewZip.Write strZipHeader
oNewZip.Close
Set oNewZip = Nothing
WScript.Sleep 5000
FolderGroup = Array(objFolder1,objFolder2,objFolder3)
FolderToZip = "FolderGroup"
ZipToFile = "C:\Win7tools\Test Script.zip"
Set ShellApp = CreateObject("Shell.Application")
Set Zip = ShellApp.NameSpace(ZipToFile)
'Set Folder = ShellApp.NameSpace(FolderToZip)
ShellApp.NameSpace(FolderGroup).CopyHere Zip.NameSpace(ZipToFile)
WScript.Sleep 10000
set ShellApp = Nothing
set FolderToZip = Nothing
set ZipToFile = Nothing