i have 3 csv files in a folder which contains eid, ename, country fields, and my 5 csv files names are test1_20120116_034512, test1_20120116_035512,test1_20120116_035
In your DataFlow task, add a Derived Column Transformation. The value of CurrentFile will be the fully qualified path to the file. As you only want the file name, I would look to use a replace function on that with the base folder and then strip the remaining slash. This does not strip the file extension but you can add yet another call to REPLACE and substitute an empty string
REPLACE(REPLACE(@[User::CurrentFile], @[User::RootFolder], ""), "\\", "")The above expects it to look like
I believe you've done something in your approach that I did not do. You should see a warning about possible truncation but given the values discussed in this and the preceding question, I don't believe the 4k limit on expressions will be of concern.
Displaying the derived column

Demonstrating the derived column does work

I will give you a +1 for providing an approach I wasn't aware of, but you'll still need to add a derived column to match your provided format (base path name)

Full path is provided from the custom properties. Use the above REPLACE section to remove the path info except use the column [FileName] instead of @[User::CurrentFile]

I tried to get the filename through the procedure which Billinkc has given, but its throwing me error stating that filename column failed becaue of truncation error..
Any how i tried different approach to load file name into table.
steps i have used
1. right click on flat file Source and click on show advanced edito for Flat file
2. select component Properties tab
3. Inside that Custom Properties section ---> it has a property FileNameColumnName
I have assigned Filename to that column property like
FileNameColumnName----> FileName thats it, am able to get the filename into my destination table..