he is right you can always use native query/js helper functions i just assumed you want to use jquery manually and $ is the same as jquery :)
On 13 Okt., 22:12, Michael T <michael.to...@gmail.com> wrote: > Tomfox, > > Just so you know, you can achieve a similar result to what EUROMARK > was suggesting using the JsHelper's "request" method. This might help > in that you have to write less raw Javascript? See the book for more > details:http://book.cakephp.org/view/1593/Methods > > Also, once you send the form data via POST, you can handle it with an > action like this in your controller: > > function handle_ajax_request() { > if($this->RequestHandler->isAjax()) { > $my_field = $this->data['my_field']; > ... > } > > } > > (Note: you need to have the RequestHandler component present in your > controller) > > Hope this helps! > > On Oct 14, 12:00 am, Tomfox Wiranata <tomfox.wiran...@gmail.com> > wrote: > > > > > first, thx to everyone.. > > > I am using 1.3 and i dont mind using POST instead of GET. seems to > > make more sense anyay.... so based on both of your answers I'll try > > > to use a form , sending with post and accept the data in my controller > > via request.. > > > @ EUROMARK: > > > so when i send the data with ajax to my controller, how should I > > receive it? with this->params? or $_POST???? > > > why do you use Jquery instead of # ? do i need to import a specific > > jquery file to make that work? i got the jquery1.4.2 already > > > thx :) > > > On 13 Okt., 11:46, euromark <dereurom...@googlemail.com> wrote: > > > > use jquery .ajax() > > > this will work > > > > example: > > > > jQuery('.delAjax').click(function(e) { > > > e.preventDefault(); > > > var targeturl = jQuery(this).attr("href"); > > > var id = jQuery(this).parent('td').parent('tr'); > > > $.ajax({ > > > type: "post", url: targeturl, > > > //beforeSend: showRequest, //show loading just > > > when link is clicked > > > //complete: showResponse, //stop showing loading > > > when the process > > > is complete > > > success: function(html){ //so, if data is > > > retrieved, store it in > > > html > > > if (html == '') { > > > id.hide(); > > > } else { > > > alert(html); > > > } > > > } > > > }); > > > }); > > > > On 13 Okt., 11:17, Michael T <michael.to...@gmail.com> wrote: > > > > > I had similar problems like yours and I found that using Firebug to > > > > debug what was being sent back to the server via Javascript was *very* > > > > helpful. If you haven't done so already, you should install it. > > > > > What version of CakePHP are you using? I know that in 1.3 you can use > > > > the JsHelper's 'request' method and you can specify that it does a > > > > POST instead of a GET. But this might only work if you employ a > > > > <form>. > > > > > Otherwise, if you want to continue using GET, can't you just get the > > > > passed value from $this->params['url']['value'] ? > > > > > Hope this helps. > > > > > On Oct 13, 9:30 pm, Tomfox Wiranata <tomfox.wiran...@gmail.com> wrote: > > > > > > but dont I need a form when I use this->data??? and still i have to > > > > > pass them in my view. but how? > > > > > > just checked....the load function is a get command, saysd > > > > > api.jquery.com > > > > > "This method is the simplest way to fetch data from the server. It is > > > > > roughly equivalent to $.get(url, data, success) " > > > > > > so i definitely have to find a different way to pass data to my > > > > > controller.....!!!! > > > > > > thanks > > > > > > On 13 Okt., 00:40, euromark <dereurom...@googlemail.com> wrote: > > > > > > > you can post it in cake style and use $this->data to access it (not > > > > > > by > > > > > > using $_POST) > > > > > > > but besides that: > > > > > > sure that "load" is a POST and not GET command? > > > > > > what does firebug say? > > > > > > > On 13 Okt., 00:25, Tomfox Wiranata <tomfox.wiran...@gmail.com> > > > > > > wrote: > > > > > > > > hmm...why not?? > > > > > > > > because i am sending data from my view to the controller? how else > > > > > > > would i process a rating? > > > > > > > > On 13 Okt., 00:21, euromark <dereurom...@googlemail.com> wrote: > > > > > > > > > its not very cake like, what you are doing there, i'm afraid > > > > > > > > > On 13 Okt., 00:00, Tomfox Wiranata <tomfox.wiran...@gmail.com> > > > > > > > > wrote: > > > > > > > > > > saveRating is supposed to call the controller function > > > > > > > > > "rate()" and > > > > > > > > > passes the rating value: > > > > > > > > > > $('#ratingt').load('rate', {data: ratingValue}); > > > > > > > > > > the action is: > > > > > > > > > > function rate() > > > > > > > > > { > > > > > > > > > $rating = $_POST['value']; > > > > > > > > > $this->set('rating', $rating); > > > > > > > > > $this->render('render_rating_count','ajax'); > > > > > > > > > } > > > > > > > > > > I am rendering a view where the result is echoed. for testing > > > > > > > > > purposes :) > > > > > > > > > > On 12 Okt., 23:50, Miles J <mileswjohn...@gmail.com> wrote: > > > > > > > > > > > Ok so whats the URL that saveRating() points to? And what > > > > > > > > > > does the > > > > > > > > > > action look like. > > > > > > > > > > > On Oct 12, 2:40 pm, Tomfox Wiranata > > > > > > > > > > <tomfox.wiran...@gmail.com> wrote: > > > > > > > > > > > > hi, > > > > > > > > > > > > when a user clicks on a product link a url like this > > > > > > > > > > > opens: products/ > > > > > > > > > > > show/45 > > > > > > > > > > > > 45 is the products id. this id is part of a select > > > > > > > > > > > statement to > > > > > > > > > > > retrieve the correct information from the database. > > > > > > > > > > > controller function: > > > > > > > > > > > > show($id=null){ > > > > > > > > > > > some code... > > > > > > > > > > > SELECT * from products WHERE products.id=$id...; > > > > > > > > > > > some code... > > > > > > > > > > > > } > > > > > > > > > > > > now the user has the opportunity to rate this product. > > > > > > > > > > > after the user > > > > > > > > > > > rated i wanna show the new result and the new count. of > > > > > > > > > > > course with > > > > > > > > > > > ajax feeling. so I use JQUERY to pass the rating value to > > > > > > > > > > > a function > > > > > > > > > > > named "rate". > > > > > > > > > > > > $("#lkbl_rating_stars_list").stars({ > > > > > > > > > > > inputType: "select", > > > > > > > > > > > cancelShow: false, > > > > > > > > > > > callback: function(value, link){ > > > > > > > > > > > saveRating(); > > > > > > > > > > > } > > > > > > > > > > > }); > > > > > > > > > > > > function saveRating() > > > > > > > > > > > { > > > > > > > > > > > var ui = > > > > > > > > > > > $("#lkbl_rating_stars_list").data("stars"); > > > > > > > > > > > var ratingValue = > > > > > > > > > > > ui.options.value; > > > > > > > > > > > > > > > > > > > > > > $('#ratingt').load('rate', {data: ratingValue}); > > > > > > > > > > > > } > > > > > > > > > > > > now cake throws error > > > > > > > > > > > "Warning 512: SQL error. cant find column rate". looking > > > > > > > > > > > at the sql > > > > > > > > > > > statement that is generated: > > > > > > > > > > > > SELECT * from products WHERE products.id=rate...; > > > > > > > > > > > > that tells me, that jquery send the data thru the URL and > > > > > > > > > > > somehow the > > > > > > > > > > > $id is replaced by "rate"...which makes no sense :) > > > > > > > > > > > > HOW do I have to pass data with jquery to my controller > > > > > > > > > > > without > > > > > > > > > > > screwing with the url? is .load not correct? > > > > > > > > > > > > appreciate it. thanks Check out the new CakePHP Questions site http://cakeqs.org and help others with their CakePHP related questions. 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