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what is the proper way to navigate away from a view before showing it?

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user1843640
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user1843640 asked on 08 Dec 2012, 07:24 PM
When an $ajax call is made in a view's show or beforeShow event, how/when/where should we navigate to another view from the $ajax success or error callback?  What I am currently doing does not work properly.

Consider the following example with 3 views:

initialView
errorView
ajaxView

  • user starts on initialView and clicks a link to navigate to #ajaxView
  • ajaxView does an $ajax call in its beforeShow event.  
  • on success, the $ajax success callback does some processing and the view is displayed
  • on error, the  error callback evaluates the error and navigates to #errorView
The above steps don't work.  errorView is not displayed.  Here is the jsfiddle for above.


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Petyo
Telerik team
answered on 10 Dec 2012, 02:32 PM
Hello,

You miss the e.preventDefault call. Check the docs on the corresponding topic. In addition to that, you have to create a synchronous ajax call.

All the best,
Petyo
the Telerik team
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user1843640
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answered on 10 Dec 2012, 07:53 PM
I'm aware of e.preventDefault but I don't have access to the event (e) from the call back.  In most cases the ajax calls need to be async so that the UI is responsive... so I need another solution.  An example use case is authentication  an async data request is made, the error callback receives the 401 status code and redirects to the login view.  I am open to suggestions.
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Petyo
Telerik team
answered on 11 Dec 2012, 07:58 AM
Hi,

In that case, you may consider handling the click event, performing the ajax call, and (if everything is correct), navigate to the respective page using app.navigate. 

Regards,
Petyo
the Telerik team
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