DSXchange

Prove your design. How many columns in the record schema, and how many fields appear in your output file? What happens if you change the Final Delimiter character (just for the test) to something else, say ";".

Thirteen.

Unless there is operator combination or some other optimization happening.

Definitely not. 50 jobs might process one row each or might process crores of rows each. You can not make that decision on something as arbitrary as simply "number of jobs".

And please note, as Craig pointed out, there is no such thing as a DataStage Scheduler

Test it with US-only data. Test it with COUNTRY rule set only. Run just the US data through USPREP rule set before applying the others. Verify that that works, then we'll investigate further.

Set the Final Delimiter property value to None.

If you have only one link in the job, you will need to put some innocuous stage (such as a Copy stage) between the two Sequential File stages.

No.

You use a DB2 Enterprise stage for that.

Job sequence.
User variables activity to convert the list to delimited list.
Start Loop activity using user variable to control list.
It's been explained before. Search.

Welcome aboard. DataStage clients have a built-in limit to how long they may wait to establish a connection. So do most internet browsers. This error usually occurs when there is some issue with the network between the client and the server - a defective modem or NIC, or port #31538 not enabled thro...

Of course there is. But the routine does not ship with the product.

There are some who could be hired to construct one for a fee.

Looks like you need to reset the tunnel connection to a longer inactivity timeout interval.

No problems in version 7.x, provided you get the NLS map correct.

I avoid user-defined SQL wherever possible, as it's one extra level of maintenance. The only time I've ever had to use it was for a UNION.

Not easy. Can you pre-process to replace them with non-breaking space ( ) characters?

That's essentially my thoughts, with the caveat given in my earlier post.

You need to partition on both key columns, so that for any value of the key [L1,D1] all corresponding values will be found on the same partition. That said, if you partition one input on [L1,D1] and one on [L1] alone, then you would still achieve the principle of key adjacency, because there are no ...