Hello All
I have the follow little question.

I have a process with 4 task, the process flows like this:


Task1 --> Task2
Task2 --> Task3
if Task3 is rejected by someone go back to Task 1
else Task3 --> Task4

the question is:
Can I go back in previous task already executed??
I Know that bpel does not support this kind of behavior (it seems like a state machine).

Any idea, thanks in advande

Victor,

you can use a looping sub-process (in the basic shapes).

Then put your Task1, Task2 and Task3 in the subprocess, and use "Task3 rejected" as the condition for looping (which will go back to Task1. Then link the looping sub-process to Task4.

Some relevant examples:
http://bpms.intalio.com/content/view/164/83/
http://bpms.intalio.com/content/view/131/83/

Thanks,
Pascal.

Hi Pascal, thanks for your response.

What about if the process grows :woohoo: . I means, if the Task 4 now relies on the Task 2??? and so on.

The fact is, there is a way to define or implement this kind of behavior without incurring in building an ad hoc solution like this.


any idea, thanks in advance!!!!!

Victor, there is no silver bullet!

The underlying execution language (BPEL) only supports nested blocks. So nesting looping sub-processes is your only option. Granted as your process grows in complexity, so will the conditions you need to express what you want, including duplicating some of the activities.

That said, in version 3, at the modeling level, we used to support looping backward (in most cases). And the tool was doing the conversion I'm suggesting above. And that's something we will be adding back beyond the 5.0 timeframe.

Note that conceptually, you can find an equivalence between the 2 solutions, but one is easier for us to understand (looping back) while the other approach is easier to implement (and that's why the BPEL specification uses nesting).

Thanks,
Pascal.

Post edited by: psq, at: 2007/02/07 08:36

Yes Pascal, you're right.

BPEL just support this kind of building blocks,

Thanks for your time and the explanation.

;)

I've created a post that explains this in more details: Looping backward

Hope this helps.

Thanks,
Pascal.