list of files in a job
Moderators: chulett, rschirm, roy
-
- Participant
- Posts: 27
- Joined: Thu Jun 26, 2008 6:13 am
- Location: chennai
list of files in a job
Can anyone provide me the unix command for getting the list of all the input and output files(including datasets and filesets) used in a job???
ramkarthik
-
- Participant
- Posts: 54607
- Joined: Wed Oct 23, 2002 10:52 pm
- Location: Sydney, Australia
- Contact:
Certainly. There isn't one.
The DataStage repository is a database so, in the same sense that there's no UNIX command for getting information out of, say, Oracle tables, there's no UNIX command for getting information out of the DataStage repository.
The DataStage repository is a database so, in the same sense that there's no UNIX command for getting information out of, say, Oracle tables, there's no UNIX command for getting information out of the DataStage repository.
IBM Software Services Group
Any contribution to this forum is my own opinion and does not necessarily reflect any position that IBM may hold.
Any contribution to this forum is my own opinion and does not necessarily reflect any position that IBM may hold.
-
- Participant
- Posts: 27
- Joined: Thu Jun 26, 2008 6:13 am
- Location: chennai
thank you
i am aware of repository in datastage (Universe database). Jobs,routines,shared containers and table defenitions are stored in repository then how will i get the list of files in a job? and file is not a job object
ramkarthik
-
- Participant
- Posts: 54607
- Joined: Wed Oct 23, 2002 10:52 pm
- Location: Sydney, Australia
- Contact:
That's the $64 million dollar question, isn't it? Further, your quest may be hampered by use of parameterized directory/folder names and/or file names. So it becomes a very complex query indeed.
Probably the easiest thing to do is to generate a job report (using dsjob -report command), and to parse the file name properties out of that. Search for the string "File name:" (for files) in the job report, and extract the remainder of each such line.
Probably the easiest thing to do is to generate a job report (using dsjob -report command), and to parse the file name properties out of that. Search for the string "File name:" (for files) in the job report, and extract the remainder of each such line.
IBM Software Services Group
Any contribution to this forum is my own opinion and does not necessarily reflect any position that IBM may hold.
Any contribution to this forum is my own opinion and does not necessarily reflect any position that IBM may hold.
-
- Participant
- Posts: 27
- Joined: Thu Jun 26, 2008 6:13 am
- Location: chennai
-
- Participant
- Posts: 54607
- Joined: Wed Oct 23, 2002 10:52 pm
- Location: Sydney, Australia
- Contact: