Magento Ajax - How to get only body part? -

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i trying use ajax call magento. when call block page via ajax, html including head, css, javascript, , body. how body part?

if can provide little more information "block page" calling may easier discern issue. default, magento includes <default> layout tag pages, give page headers , footers on ajax calls.

to send page without extra, have few options. firstly, can simple set output manually on own, avoiding layout system entirely. magento one-page checkout feature:

$result = array( 'foo' => 'foo', 'bar' => 'bar', ); $this->getresponse()->setbody(zend_json::encode($result));

you can modify method use custom layout handler this:

protected function loadpage() {     $layout = $this->getlayout();     $update = $layout->getupdate();     $update->load('your_custom_handle');     $layout->generatexml();     $layout->generateblocks();     $output = $layout->getoutput();      $result = array( 'outputhtml' => $output, 'othervar' => 'foo', );     $this->getresponse()->setbody(zend_json::encode($result));         }

and in layout file:

<your_custom_handle>     <remove name="right"/>     <remove name="left"/>      <block type="module/block" name="root" output="tohtml" template="module/template.phtml"/> </your_custom_handle>

a second option, if want use layouts, define alternative default layout. when call $this->loadlayout(); in magento controllers, can specify handle other <default> descend from. example magento product controller be:

$this->loadlayout('popup');

this layout defined default within main.xml layout file, , renders popup.phtml template, , may suitable use.

if still have trouble, let me know , can try other things. hope helps.

thanks, joe


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