java - Disabling broadcast in SimpMessageTemplate.sentToUser(...) in Spring 4.2.4 -


i trying connect in 2 sessions same spring security user , subscribe same queue using spring 4.2.4. separate messages each session. it's easy using @sendtouser, seems it's not trivial when using simpmessagingtemplate.sendtouser(...) - every session receives messages specific other session.

example @sendtouser:

@controller public class testcontroller {      @messagemapping("/queue/hello")     @sendtouser(value = "/queue/hello", broadcast = false)     public string test() {         return integer.tostring(new random().nextint());     }  } 

example simpmessagingtemplate:

template.convertandsendtouser("sessionuser{id=123}", "/queue/hello", "test"); 

i've tried adding sessionid in headers suggested example in thread: sending-error-message-in-spring-websockets unfortunately not in version of spring. have ideas?

seems setting session id in headers still works, has simpsessionid, not sessionid. assumptions problem lies wrong. summarize, can use overloaded method:

simpmessagingtemplate.convertandsendtouser("sessionuser{id=123}", "/queue/hello", "test", headers); 

where headers created by:

private messageheaders createheaders(string sessionid) {     simpmessageheaderaccessor headeraccessor = simpmessageheaderaccessor.create(simpmessagetype.message);     headeraccessor.setsessionid(sessionid);     headeraccessor.setleavemutable(true);     return headeraccessor.getmessageheaders(); } 

and access simpsessionid should use:

simpmessageheaderaccessor.getsessionid(message.getheaders()) 

hope useful someone.


Comments

Popular posts from this blog

ios - RestKit 0.20 — CoreData: error: Failed to call designated initializer on NSManagedObject class (again) -

java - Digest auth with Spring Security using javaconfig -

laravel - PDOException in Connector.php line 55: SQLSTATE[HY000] [1045] Access denied for user 'root'@'localhost' (using password: YES) -