Ok, if your page is loading normaly, then pass the information to the view/layout and in there, make the decision to activate or not, the ajax request to the bar action. How you are going to do it is up to you, I have no idea :) of what information you use and how you have chosen to implement your solution to your web site! Enjoy, John
On Mar 24, 11:32 am, alex bailey <v...@planet-generals.de> wrote: > Hi John, > > thanks for your answer could you explain a bit further how to retrieve > the condition and return it with the result? I've worked with php a > long time now but i just recently delved into ajax so I still have to > find my way around. And btw my normal page loading works without ajax > I only want a action which reacts to specified events (like a change > in the database) to be executed over and over again and "do something" > if that certain event happens. > > Regards > > Alex > > On 24 Mrz., 08:09, John Andersen <j.andersen...@gmail.com> wrote: > > > I am not sure how your page loading is working - with ajax or without > > - so here is what I would do based on your information: > > > In the first ajax call to the bar action, retrieve the dynamic > > condition and return it together with the ajax result. Then act upon > > it as you want. > > > Enjoy, > > John Check out the new CakePHP Questions site http://cakeqs.org and help others with their CakePHP related questions. You received this message because you are subscribed to the Google Groups "CakePHP" group. To post to this group, send email to cake-php@googlegroups.com To unsubscribe from this group, send email to cake-php+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/cake-php?hl=en To unsubscribe from this group, send email to cake-php+unsubscribegooglegroups.com or reply to this email with the words "REMOVE ME" as the subject.