| Newbie | | Join Date: Mar 2008
Posts: 1
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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(zipped) 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. -
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Public Class clsZip
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Public Sub CreateZipFile(ByVal sPath As String)
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Dim fos As java.io.FileOutputStream
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Dim zos As java.util.zip.ZipOutputStream
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Dim di As System.IO.DirectoryInfo
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'check , is it a file existing in this path, if true then zip a file
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If System.IO.File.Exists(sPath) Then
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Dim fInfo As New FileInfo(sPath)
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'create a zip file with same name in the same path
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fos = New java.io.FileOutputStream(sPath.Replace(fInfo.Exten sion, ".zip"))
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zos = New java.util.zip.ZipOutputStream(fos)
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'procedure to zip one File
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ZipOneFile(fos, zos, sPath)
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'check , is it a directory existing in this path,if true then zip a directory
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ElseIf System.IO.Directory.Exists(sPath) Then
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'create a zip file with same name in the same path
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fos = New java.io.FileOutputStream(sPath & ".zip")
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zos = New java.util.zip.ZipOutputStream(fos)
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di = New System.IO.DirectoryInfo(sPath)
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'procedure to zip a directory
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ZipDirectory(fos, zos, di, sPath)
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End If
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zos.close()
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fos.close()
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zos.flush()
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fos.flush()
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End Sub
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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)
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Dim fis As java.io.FileInputStream
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Dim ze As java.util.zip.ZipEntry
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'to get file info from the directory
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Dim fInfos As System.IO.FileInfo() = di.GetFiles
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Dim fInfo As System.IO.FileInfo
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For Each fInfo In fInfos
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'give the zip entry or the folder arrangement for the file
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ze = New java.util.zip.ZipEntry(fInfo.FullName.Substring(SR ootDir.LastIndexOf("\")))
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'The DEFLATED method is the one of the methods to zip a file
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ze.setMethod(ze.DEFLATED)
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zos.putNextEntry(ze)
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'Input stream for the file to zip
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fis = New java.io.FileInputStream(fInfo.FullName)
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'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)
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CopyStream(fis, zos)
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zos.closeEntry()
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fis.close()
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Next
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'If the directory contains the sub directory the call the same procedure
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Dim dinfos As System.IO.DirectoryInfo() = di.GetDirectories()
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Dim dinfo As System.IO.DirectoryInfo
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For Each dinfo In dinfos
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ZipDirectory(fos, zos, dinfo, SRootDir)
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Next
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End Sub
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Private Sub ZipOneFile(ByVal fos As java.io.FileOutputStream, ByVal zos As java.util.zip.ZipOutputStream, ByVal sFullName As String)
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Dim fis As java.io.FileInputStream
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Dim ze As java.util.zip.ZipEntry
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'give the zip entry or the folder arrangement for the file
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ze = New java.util.zip.ZipEntry(sFullName.Substring(sFullNa me.LastIndexOf("\")))
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'The DEFLATED method is the one of the methods to zip a file
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ze.setMethod(ze.DEFLATED)
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zos.putNextEntry(ze)
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'Input stream for the file to zip
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fis = New java.io.FileInputStream(sFullName)
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'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)
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CopyStream(fis, zos)
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zos.closeEntry()
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fis.close()
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End Sub
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Private Sub CopyStream(ByVal src As java.io.FileInputStream, ByVal dest As java.util.zip.ZipOutputStream)
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Dim reader As New java.io.InputStreamReader(src)
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Dim writer As New java.io.OutputStreamWriter(dest)
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While reader.ready
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writer.write(reader.read)
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End While
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writer.flush()
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End Sub
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regards
Nagashree
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