One file regarding the info of the all files in the folder

[parvathi]

Hi
I have a requriment like this i have job which generates a five .csv extension files.

Once the files are generated i have to create a file such that each row in the file should contain the file name and the noof records in that file
and the next row the other file nameand records of that file..
This process should be done for all the files with the .csv extension

Is it better to do with the datastage server job or or run a unix shell script to get the above info

[ArndW]

I think that this is a clear case of using a UNIX shell script. The KISS principle applies and a short script is better than a medium-sized DS job/program.