Wednesday, October 15, 2008

Zip and unzip files in a folder

Many a time we need to zip or unzip more then one file present in a folder. SSIS provides a easy way to handle this.

For my example I am using 7-zip though we have many other compress software like (pkzip or winzip)

Steps to zip/unzip files
A) Declare a variable FileName as string datatype


B) Use ForEach loop to loop through the files in the folder and map file names to FileName variable
a) Provide folder name inEnumerator configuration


b) map FileName Variable


C) Use Execute Process task to zip/unzip to individual files
a) Browse to Exe of 7z in Executable


b) Use expression to configure the argument


D) Execute the package


Use the files........
at
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21 comments:

  1. AnonymousAugust 5, 2009 at 1:49 PM

    Dear Rahul,

    Thank you for posting this topic, very very helpfull to see a step-by-step guide with screenshots.

    Namasté,

    Gahan Zwart

    ReplyDelete
  2. Jeffery WilliamsMay 22, 2010 at 1:21 AM

    I am trying to do this in a stored procedure. The 7zip program is executing but the return is that it can not find the archive. I have a print statement to see what I am passing to the OS and if I copy / paste it my command line works. But does not find the file from SQL execution. Any ideas anyone.

    C:\7zip\7zip\7z.exe e C:\Users\jwilliams\Documents\EConnect\FirstMerrit\SourceFiles\810571sn_4.zip pP810571

    The above it the result of running my SQL script.

    HEre is the actual SQL script:

    ALTER PROCEDURE UnZipFile @CustomerID int

    AS

    declare @ZipFileName VarChar(max)
    declare @SQLCommand VarChar(max)
    DECLARE @DataZipPass as varchar(10)


    SELECT @DataZipPass = DataZipPass
    FROM Customers
    WHERE Customers.CustomerID = @CustomerID


    SET @ZipFileName = 'C:\Users\jwilliams\Documents\EConnect\FirstMerrit\SourceFiles\810571sn_4.zip'


    SET @SQLCommand =
    'exec master..xp_cmdshell ' + '''' + 'C:\7zip\7zip\7z.exe e '
    + @ZipFileName + ' p' +@DataZipPass + ' -y ' + ''''

    PRINT 'Extraction Command sent: ' + cast(@SQLCommand as varchar(max))
    PRINT ''

    EXEC (@SQLCommand)

    ReplyDelete
  3. ashish jindalMarch 18, 2011 at 3:58 AM

    This comment has been removed by the author.

    ReplyDelete
  4. ashish jindalMarch 18, 2011 at 4:00 AM

    Hello, it was a really good example. it helped me in completing Unzip file task. i have a small query...
    i need the name of output file.
    i provided param StandardOutputVariable = @[User::NewFileName]
    but i can't capture the filename.
    am i missing something

    ReplyDelete
  5. Rahul KumarMarch 18, 2011 at 10:33 AM

    @ashish: to get output file name depends on the utility you would use.
    StardardOutputVariable will get all the information return from the process,if the utility returns the just the file name..you are gud but usually Utilities return lot of information back like success status, size etc and file name is just one of them.

    ReplyDelete
  6. AnonymousJune 23, 2011 at 2:26 AM

    I getting the following error: The process exit code was "7" while the expected was "0"

    Can you please help me how I can fix this error.


    Thanks

    ReplyDelete
  7. Rahul KumarJune 23, 2011 at 10:48 AM

    @ "I getting the following error: The process exit code was "7" while the expected was "0"

    Can you please help me how I can fix this error"

    There must be some error, might be in command stmt.. could you get the stmt and run it in command prompt or paste here so that i can look at it.

    ReplyDelete
  8. shahbazNovember 28, 2011 at 12:48 PM

    THANKS FOR EXCELLENT WORK RAHUL. I APPLIED TO MY TEST AND IT WORKS LIKE CHARM.

    ReplyDelete
  9. Rahul KumarNovember 28, 2011 at 2:25 PM

    @shahbaz...pleasure!!

    ReplyDelete
  10. TP5April 11, 2012 at 12:58 AM

    Hi...I'm also getting the prcess exit code was 7 - please help

    ReplyDelete
  11. TP5April 11, 2012 at 1:00 AM

    I'm also getting the error:the process exit code was "7" while the expected was "0" - please help

    ReplyDelete
  12. David RoodmanApril 18, 2012 at 12:15 AM

    Thanks, this was just what I needed. It solved my problem and gave me a crash course in programming integration services packages.

    ReplyDelete
  13. UnknownAugust 14, 2012 at 8:22 PM

    Hello Rahul,

    Can you please explain your arguments used?

    ReplyDelete
  14. user1December 21, 2012 at 9:44 PM

    Iam getting below error ,Please help
    The process exit code was "7" while the expected was "0".

    ReplyDelete
  15. AnonymousFebruary 12, 2013 at 3:55 AM

    Hi, I tried to unzip a file using 7zip at command line with the following command:

    C:\Program Files\7-Zip>7z e c:\easy.zip -oc:\new folder\

    I got the following error:
    No files to process

    Files: 0
    Size: 0
    Compressed: 4790659

    C:\Program Files\7-Zip>

    Please help.

    What am I doing wrong?

    ReplyDelete
  16. NiallMay 21, 2014 at 6:36 PM

    Hurray, thanks! Worked first time.

    ReplyDelete
  17. AnonymousFebruary 9, 2015 at 6:02 AM

    I appreciate your tutorial it helped me a lot! Thank You!!
    Jeff

    ReplyDelete
  18. AnonymousJune 2, 2015 at 2:38 AM

    what is @[User::NewFileName]

    ReplyDelete
  19. UnknownSeptember 22, 2015 at 8:09 AM

    Age is an issue of mind over matter. If you don't mind, it doesn't matter. See the link below for more info.


    #issue
    www.ufgop.org

    ReplyDelete
  20. UnknownMay 17, 2016 at 8:25 PM

    Hi Rahul,

    Could you please tell me What 'e -oc:/rahul' is in Argument??

    ReplyDelete
  21. UnknownJanuary 6, 2018 at 11:15 PM

    You have made nice post but you can know more about 7-zip (compressor/extractor) it at:
    https://basusoftgame.blogspot.com/2017/12/7-Zip-Unzip-file-for-Windows.html

    ReplyDelete

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