SSIS Expressions Problem

Question

Hi All,
Please help on the SSIS issue.
i have a package.It contains Foreach loop container,Inside foreach loop container,I have one Dataflow task(Inside dataflow task,i am using Oledb source(table),I am populating 
Data into Flat file)with flat file name YYYYMMDD_HH24MISS.txt. After i want to Zip that file with same file name. 
YYYYMMDD_HH24MISS.txt.zip,After that i want copy that Zip file into another location.
Thanks in advance...

Answer 1

Hi BADDULAS,
According to the screenshot in your initial post, you are using an Execute Process Task to compress the .txt file generated by the Data Flow Task. Let’s assume that you are using the common free software 7-Zip as the executable of the Execute Process Task,
then you can refer to the following blog:
http://sqlage.blogspot.com/2013/12/ssis-how-to-compress-and-archive-file.html 
To get the expected name for the txt file, you can create another String type SSIS variable @[User::FileName] using the following expression (set EvaluateAsExpression property of the variable to True):
REPLACE(SUBSTRING((DT_STR,30,1252)GETDATE(),1,10),"-","")+"_"+ SUBSTRING((DT_STR,30,1252)GETDATE(),12,2) + "24MISS.txt"
The FileName variable should be used by the ConnectionString property of the Flat File Connection Manager that is used by the Flat File Destination in the Data Flow Task.
Besides, the variable FileName should also be used in the expression of the VarSourcePath variable and the VarArchivePath variable as follows:
VarSourcePath:  “C:\\Temp\\Source\\” + @[User::FileName]
VarArchivePath: “C:\\Temp\\Destination\\” + REPLACE(@[User::FileName], ”.txt”, “”)
Regards,
Mike Yin
TechNet Community Support