i am new to ZF i want to create ajax link that will go to "task" controller and "ajax" action do something like this
$registry = Zend_Registry::getInstance();
$DB = $registry['DB'];
$sql = "SELECT * FROM task ORDER BY task_name ASC";
$result = $DB->fetchAll($sql);
than put the result in this div
<div id="container">container</div>
this is my view where i am doing this
<?php echo $this->jQuery()->enable(); ?>
<?php echo $this->jQuery()->uiEnable(); ?>
<div id="container">container</div>
<?php
echo $this->ajaxLink("Bring All Task","task/ajax",array('update' => '#container'));
?>
i dont know the syntax how i will do this , retouch my code if i am wrong i searched alot but all in vain plz explain me thanking you all in anticipation also refer me some nice links of zendx_jquery tutorial
This should work:
and in your ajax.phtml
regarding db you have to setup it first somewhere earlier in your code, for example front controller index.php or bootstrap.php, for example: