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SSIS Unpack a ZIP file with the Script Task

A while ago I needed to unpack a couple of zip files from SSIS. There is no Microsoft

SSIS task that contains this functionality so I searched the Internet. It seems that there

are quite some third party tools that offer this functionally. It's also possible to download

custom SSIS tasks. I personally always try to avoid third party tools and custom tasks so

I searched on.It seemed there is a way to unzip files from SSIS with the Script Task. With some Visual

Basic code using the Visual J# Library you can do the job. In this blog post I will use a

Foreach Loop Container to loop through a folder that contains multiple zip files and unzip

them one-by-one.

Make sure you have the Microsoft Visual J# Redistributable Package installed because a

reference to vjslib.dll (Visual J# Library) is needed in the Script Task. Download it here

for free.

Drag and drop a Foreach Loop Container on the Control Flow and create three variables

with scope on the Foreach Loop Container:

Now configure the Foreach Loop Container:

- Enumerator: Foreach File Enumerator

- Files: *.zip

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- Retrieve file name: Name and extension

Next click on the + next to Expressions add the following expression to connect the

SourceFolder variable to the Directory property of the Foreach Loop Container:

Now go to the Variable Mappings and select the FileName variable on Index 0. Doing this

we will be able to access the current file name when the Foreach Loop Container

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enumerates the zip files.

Now drag and drop a Script Task on the Control Flow, inside the Foreach Loop Container:

Open the Script Task Editor and do the following:- Set the ScripLanguage on: Microsoft Visual Basic 2008

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- Select our three ReadOnlyVariables using the new SSIS2008 Select Variables window:

Now click Edit Script and copy/paste the following script: Imports SystemImports System.DataImports System.MathImports Microsoft.SqlServer.Dts.RuntimeImports java.util.zip

PublicSub Main()Try

Dim strSourceFile AsStringDim strDestinationDirectory As String

'MsgBox("Current File: " & Dts.Variables("FileName").Value.ToString)strDestinationDirectory = Dts.Variables("DestinationFolder").Value.ToStringstrSourceFile = Dts.Variables("SourceFolder").Value.ToString &

Dts.Variables("FileName").Value.ToStringDim oFileInputStream As New java.io.FileInputStream(strSourceFile)Dim oZipInputStream As New java.util.zip.ZipInputStream(oFileInputStream)Dim bTrue As Boolean = TrueDim sbBuf(1024) As SByteWhile 1 = 1

Dim oZipEntry As ZipEntry = oZipInputStream.getNextEntry()If oZipEntry IsNothingThenExitWhileIf oZipEntry.isDirectory Then

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If NotMy.Computer.FileSystem.DirectoryExists(strDestinationDirectory &oZipEntry.getName) Then

My.Computer.FileSystem.CreateDirectory(strDestinationDirectory &oZipEntry.getName)

End IfElse

Dim oFileOutputStream As Newjava.io.FileOutputStream(strDestinationDirectory.Replace("\", "/") & oZipEntry.getName())

While 1 = 1

Dim iLen As Integer = oZipInputStream.read(sbBuf)If iLen < 0 Then Exit WhileoFileOutputStream.write(sbBuf, 0, iLen)

End WhileoFileOutputStream.close()

End IfEnd WhileoZipInputStream.close()oFileInputStream.close()

Catch ex As ExceptionThrow New Exception(ex.Message)

End TryEnd Sub

End Class

Now only one thing needs to be done, add a reference to vjslib.dll (Visual J# Lib rary):

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&

Your unzip solution is ready now! For testing purposes you can uncomment the following

line in the script to see the file name of each processed zip file in a message box at

runtime:

'MsgBox("Current File: " & Dts.Variables("FileName").Value.ToString)