[jboss-user] [JBoss jBPM] - A question about jBPM process execution.

youhaodeyi do-not-reply at jboss.com
Tue May 13 09:23:38 EDT 2008


This is my code:

  | 		ProcessDefinition processDefinition = ProcessDefinition.parseXmlInputStream(new FileInputStream(
  | 					"jbpm/helloworld/processdefinition.xml"));
  | 			
  | 			ProcessInstance processInstance = new ProcessInstance(processDefinition);
  | 
  | 			Token token = processInstance.getRootToken();
  | 			
  | 			token.signal();
  | 			System.out.println(token.getNode());
  | 			
  | 			token.signal();
  | 			System.out.println(token.getNode());
  | 
This is the processdefinition.xml:
<process-definition  xmlns=""  name="helloworld">


	<start-state name="start">
		
	</start-state>


	
		
	


	<end-state name="end"></end-state>


</process-definition>

My question is that why I need invoke signal two times. I assume that I only need to signal the first state and it will transition to the second state and then to the end state. The transition has been specified in the configuration file.

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