Decompressing the file or directory using VB.NET class

[nagashree]

Hi,

i want to compress the file or directory i am able to do that can anybody help me to decompress the file when i double click on the zipped file i am getting the error message as compressed(zipp ed) folder is invalid or corrupted. I am using VB.NET .Please help me out in this regard...Thank you in advance.

Here is the code for that.

Public Class clsZip

Public Sub CreateZipFile(ByVal sPath As String)
Dim fos As java.io.FileOutputStream
Dim zos As java.util.zip.ZipOutputStream
Dim di As System.IO.DirectoryInfo

'check , is it a file existing in this path, if true then zip a file
If System.IO.File.Exists(sPath) Then
Dim fInfo As New FileInfo(sPath)

'create a zip file with same name in the same path
fos = New java.io.FileOutputStream(sPath.Replace(fInfo.Exten sion, ".zip"))
zos = New java.util.zip.ZipOutputStream(fos)

'procedure to zip one File
ZipOneFile(fos, zos, sPath)

'check , is it a directory existing in this path,if true then zip a directory
ElseIf System.IO.Directory.Exists(sPath) Then

'create a zip file with same name in the same path
fos = New java.io.FileOutputStream(sPath & ".zip")
zos = New java.util.zip.ZipOutputStream(fos)
di = New System.IO.DirectoryInfo(sPath)

'procedure to zip a directory
ZipDirectory(fos, zos, di, sPath)
End If


zos.close()
fos.close()
zos.flush()
fos.flush()
End Sub

Private Sub ZipDirectory(ByVal fos As java.io.FileOutputStream, ByVal zos As java.util.zip.ZipOutputStream, ByVal di As System.IO.DirectoryInfo, ByVal SRootDir As String)
Dim fis As java.io.FileInputStream
Dim ze As java.util.zip.ZipEntry

'to get file info from the directory
Dim fInfos As System.IO.FileInfo() = di.GetFiles
Dim fInfo As System.IO.FileInfo

For Each fInfo In fInfos
'give the zip entry or the folder arrangement for the file
ze = New java.util.zip.ZipEntry(fInfo.FullName.Substring(SR ootDir.LastIndexOf("\")))

'The DEFLATED method is the one of the methods to zip a file
ze.setMethod(ze.DEFLATED)
zos.putNextEntry(ze)

'Input stream for the file to zip
fis = New java.io.FileInputStream(fInfo.FullName)

'Copy stream is a simple method to read a file input stream (file to zip) and write it to a file output stream(new zip file)
CopyStream(fis, zos)

zos.closeEntry()
fis.close()
Next

'If the directory contains the sub directory the call the same procedure
Dim dinfos As System.IO.DirectoryInfo() = di.GetDirectories()
Dim dinfo As System.IO.DirectoryInfo
For Each dinfo In dinfos
ZipDirectory(fos, zos, dinfo, SRootDir)
Next

End Sub

Private Sub ZipOneFile(ByVal fos As java.io.FileOutputStream, ByVal zos As java.util.zip.ZipOutputStream, ByVal sFullName As String)
Dim fis As java.io.FileInputStream
Dim ze As java.util.zip.ZipEntry
'give the zip entry or the folder arrangement for the file
ze = New java.util.zip.ZipEntry(sFullName.Substring(sFullNa me.LastIndexOf("\")))
'The DEFLATED method is the one of the methods to zip a file
ze.setMethod(ze.DEFLATED)
zos.putNextEntry(ze)
'Input stream for the file to zip
fis = New java.io.FileInputStream(sFullName)
'Copy stream is a simple method to read a file input stream (file to zip) and write it to a file output stream(new zip file)
CopyStream(fis, zos)
zos.closeEntry()
fis.close()
End Sub

Private Sub CopyStream(ByVal src As java.io.FileInputStream, ByVal dest As java.util.zip.ZipOutputStream)
Dim reader As New java.io.InputStreamReader(src)
Dim writer As New java.io.OutputStreamWriter(dest)
While reader.ready
writer.write(reader.read)
End While
writer.flush()
End Sub

regards
Nagashree