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It would be nice to have a function that leaves the current route and goes back up to the parent route using all the same parameters.
Like in the router, if I have afterModel(model, transition) , I may want to immediately leave the route after the route has done something in particular but it's a bit contrived to have to spell out the full route path in this.transitionTo('route1.route2.route3') etc
The code would be more reusable if you could just say:"this.leaveCurrentRoute() or just this.exitRoute() or this.leave()", blind to what route it actually is.
Thanks.
The text was updated successfully, but these errors were encountered:
Benjy1979
changed the title
A function Route() to leave current route.
A function in the Route to just leave current route, whatever it is.
Jun 14, 2018
Hi,
It would be nice to have a function that leaves the current route and goes back up to the parent route using all the same parameters.
Like in the router, if I have afterModel(model, transition) , I may want to immediately leave the route after the route has done something in particular but it's a bit contrived to have to spell out the full route path in this.transitionTo('route1.route2.route3') etc
The code would be more reusable if you could just say:"this.leaveCurrentRoute() or just this.exitRoute() or this.leave()", blind to what route it actually is.
Thanks.
The text was updated successfully, but these errors were encountered: