Reputation: 212
I want to render output of php page into html div.That means in php page has lot of jquery stuff.suppose i used current jquery ajax to load a php page.It missed to load jquery document ready stuff.
How can i do that?
Upvotes: 1
Views: 11919
Reputation: 8476
you can use ajax observe
method
<?php echo $form->create( 'Post' ); ?>
<?php $titles = array( 1 => 'Tom', 2 => 'Dick', 3 => 'Harry' ); ?>
<?php echo $form->input( 'title', array( 'options' => $titles ) ) ?>
<label for="something">This checkbox will be send (and it will trigger Ajax reqest) </label>
<input id="something" type="checkbox" name='data[Post][something]' value='1' />
</form>
<?php
echo $ajax->observeForm( 'PostAddForm',
array(
'url' => array( 'action' => 'edit' ),
'complete' => 'alert(request.responseText)'
)
); ?>
Upvotes: 0
Reputation: 5818
Here's is the short example. jQuery API Doc
html
<div id="content"></div>
<input type="button" id="btnLoad" value="Load" />
jquery
$("#btnLoad").click(function(){
$.ajax({
type: 'POST',
url: 'page1.php',
success: function(data){
if(data != null) $("#content").text(data)
}
});
});
page1.php
<?php
echo "This is the sample data to be printed for request from AJAX";
?>
Upvotes: 0
Reputation: 2163
If you want to put the output of a php page to an html page using jquery and ajax. You may do this.
$(document).ready(function(){
$("#div1").load("demo_test.php");
});
div1 is the div which is to be updated with the php content. This link may help you.
http://www.w3schools.com/jquery/jquery_ajax_intro.asp
Upvotes: 5