Re: [Wonder-disc] How to escape from an Ajax Request without use of Direct Actions

[Amedeo M. <ame...@in...>]

ubable to get both solutions working

i'll explain the problem better...

the redirect i need is inside the haldleException method

@Override
	public WOResponse handleException(Exception _exception, WOContext _context) {
		
		CFError erp = (CFError)pageWithName("CFError", _context);
		erp.setTrace( _exception.getStackTrace() );
		erp.setMessage( _exception.getMessage() );

		WOResponse response = erp.generateResponse();

		if (AjaxUtils.isAjaxRequest(_context.request() )) {
			AjaxUtils.redirectTo(erp);
		}

		return response;
	}


but I still receive the error page rendered inside the AjaxUpdateContainer

Regards
Amedeo


On 27/gen/2010, at 18.30, Chuck Hill wrote:

> 
> On Jan 27, 2010, at 5:22 AM, Amedeo Mantica wrote:
> 
>> Hello,
>> 
>> I have an Ajax action on my page, and works fine, but under some ecircustances, i need to get out from ajax request, so if the user click the action link, i get a new page instead of ajax response....
>> 
>> the easy way for me was use AjaxUtils.isAjaxRequest then generate a response with a <script>document.location.href='myDirectAction';</script>" and this works fine...
>> 
>> but how if I want to get a component action page and not a direct action ??
>> 
>> Regards
>> Amedeo
> 
> Use AjaxUtils.redirectTo(WOComponent component)
> 
> 
> Chuck
> 
> -- 
> Chuck Hill             Senior Consultant / VP Development
> 
> Practical WebObjects - for developers who want to increase their overall knowledge of WebObjects or who are trying to solve specific problems.
> http://www.global-village.net/products/practical_webobjects
> 
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