[rules-users] Rule Flow: Diverge/Converge Issue with one-to-many paths being exected

2011-11-09 Thread dcrissman
Drools Version: 5.1 (We are looking into upgrading, but not an option at this
time)

I have need of a rule flow that will allow a diverge on one-to-many paths
and then converge and continue on only once the paths that were executed
have all finish.

Say I have path A, B,  C. The constraints on the diverge node indicate that
A  C should be executed simultaneously, but B should not. Then the converge
node should wait for A and C to finish, but not care about B as it was not
executed by the diverge.

Initial attempt was to use Diverge(OR) and Converge(AND). The problem is
that the Converge (I think) waits on B indefinitely and so the process exits
out with no error message.

It looks like n-of-m is kind of what I want, except I have no way to know
how many paths the Diverge kicked off.

Any thoughts on how this could be done?

Much Appreciated!

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Re: [rules-users] Rule Flow: Diverge/Converge Issue with one-to-many paths being exected

2011-11-09 Thread Swindells, Thomas
Can you split it into two separate diverges?
Diverge to B first and always follow the other path which then does the A  C,
Similarly at the other end wait for A C and then a second wait to add in the 
condition for B.

Thomas

 -Original Message-
 From: rules-users-boun...@lists.jboss.org [mailto:rules-users-
 boun...@lists.jboss.org] On Behalf Of dcrissman
 Sent: 09 November 2011 15:05
 To: rules-users@lists.jboss.org
 Subject: [rules-users] Rule Flow: Diverge/Converge Issue with one-to-many
 paths being exected

 Drools Version: 5.1 (We are looking into upgrading, but not an option at this
 time)

 I have need of a rule flow that will allow a diverge on one-to-many paths and
 then converge and continue on only once the paths that were executed
 have all finish.

 Say I have path A, B,  C. The constraints on the diverge node indicate that A
  C should be executed simultaneously, but B should not. Then the converge
 node should wait for A and C to finish, but not care about B as it was not
 executed by the diverge.

 Initial attempt was to use Diverge(OR) and Converge(AND). The problem is
 that the Converge (I think) waits on B indefinitely and so the process exits
 out with no error message.

 It looks like n-of-m is kind of what I want, except I have no way to know how
 many paths the Diverge kicked off.

 Any thoughts on how this could be done?

 Much Appreciated!

 --
 View this message in context: http://drools.46999.n3.nabble.com/Rule-Flow-
 Diverge-Converge-Issue-with-one-to-many-paths-being-exected-
 tp3493775p3493775.html
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Re: [rules-users] Rule Flow: Diverge/Converge Issue with one-to-many paths being exected

2011-11-09 Thread dcrissman

Swindells, Thomas wrote:
 
 Can you split it into two separate diverges?
 Diverge to B first and always follow the other path which then does the A
  C,
 Similarly at the other end wait for A C and then a second wait to add in
 the condition for B.
 
 Thomas
 

I am not sure I entirely follow your suggestion. A, B,  C could fire in any
combination, but always at least one.

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Re: [rules-users] Rule Flow: Diverge/Converge Issue with one-to-many paths being exected

2011-11-09 Thread FrankVhh
Hi,

I do not reallt know whether there is a native way in jBPM to implement the
m out of n-case you are describing. Anyway, imho, the easiest way to
roundtrip this is probably this:

Create a series of simple decision gateways: If(A has to execute), fire A;
else do nothing; If (B has to execute); fire B; else do nothing;...
Each time a task is not to be execute, the token will follow an empty
path. You can use the or convergence to join the branches together again and
lead them to the next divergence.

Hope this helps.

Regards,
Frank 


dcrissman wrote:
 
 
 Swindells, Thomas wrote:
 
 Can you split it into two separate diverges?
 Diverge to B first and always follow the other path which then does the A
  C,
 Similarly at the other end wait for A C and then a second wait to add in
 the condition for B.
 
 Thomas
 
 
 I am not sure I entirely follow your suggestion. A, B,  C could fire in
 any combination, but always at least one.
 


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