The Letter E Posted April 8, 2011 Share Posted April 8, 2011 I'm just learning the jquery ajax and still have a lot of questions, mainly because at this point it's not returning any results. I was hoping someone could point out where i'm going wrong here. Here's the main page: <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Ajax - PHP example</title> <link href="http://ajax.googleapis.com/ajax/libs/jqueryui/1.8/themes/base/jquery-ui.css" rel="stylesheet" type="text/css"/> <script src="http://ajax.googleapis.com/ajax/libs/jquery/1.5/jquery.min.js"></script> <script src="http://ajax.googleapis.com/ajax/libs/jqueryui/1.8/jquery-ui.min.js"></script> <script type="text/javascript"> $(document).ready(function() { $("#work").click(function() { $.getJSON('remote.php?ajax=1&test=12', function(data) { $("#element").html(data.value); }); }); }); </script> </head> <body> <div id="element"></div> <a href="#" id="work">DO WORK!</a> </body> </html> and here's the php it's querying: <?php if(isset($_GET['ajax']) && $_GET['ajax'] == 1){ $test = $_GET['test']; $array = array(12, 13, 34, 2, 18); $output = array(); foreach($array as $check){ $output[] .= $test * $check.'<br>'; } } if(!empty($output)){ echo json_encode($output); }else{ echo 0; } ?> This is just a test example to get my feet wet, and any help is appreciated. Thanks, E Quote Link to comment Share on other sites More sharing options...
markjoe Posted April 13, 2011 Share Posted April 13, 2011 Still unsolved? Your php script is returning a JSON array but your javascript is looking for an object with a "value" property. Changing $("#element").html(data.value); to $("#element").html(data[0]); will put "144" in the div. Hopefully this will get you on the right track. Quote Link to comment Share on other sites More sharing options...
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