PHP和JavaScript POST

我有以下代码：PHP和JavaScript POST

<html> 
<head> 
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.5.1/jquery.min.js" type="text/javascript"> 
</script> 
<script type="text/javascript"> 


function displayTweet(){ 
var i = 0; 
var limit = $("#twitter-results > div").size(); 
var myInterval = window.setInterval(function() { 
var element = $("#twitter-results div:last-child"); 
$("#twitter-results").prepend(element); 
element.fadeIn("slow"); 
i++; 
if(i==limit){ 
window.setTimeout(function() { 
clearInterval(myInterval); 
}); 
} 
},2000); 
} 

$("form#twittersearch").submit(function() { 
twitterq = $('#twitterq').attr('value'); 
$.ajax({ 
type: "POST", 
url: "search.php", 
cache: false, 
data: "twitterq="+ twitterq, 
success: function(html){ 
$("#twitter-results").html(html); 
displayTweet(); 
} 
}); 
return false; 
}); 
}); 

</script> 

</head> 
<div class="twitter_container"> 
<form id="twittersearch" method="post" action=""> 
<input name="twitterq" type="text" id="twitterq" /> 
<button type="submit">Search</button> 
</form> 
<div id="twitter-results"></div> 
</div> 
</html> 


/***************THIS IS search.php***************************/ 

<?php 
include('twitterapi.php'); 
if($_POST['twitterq']){ 
$twitter_query = $_POST['twitterq']; 
$search = new TwitterSearch($twitter_query); 
$results = $search->results(); 

foreach($results as $result){ 
    echo '<div class="twitter_status">'; 
    echo '<img src="'.$result->profile_image_url.'" class="twitter_image">'; 
    $text_n = toLink($result->text); 
    echo $text_n; 
    echo '<div class="twitter_small">'; 
    echo '<strong>From:</strong> <a href="http://www.twitter.com/'.$result->from_user.'">'.$result->from_user.'</a&glt;: '; 
    echo '<strong>at:</strong> '.$result->created_at; 
    echo '</div>'; 
    echo '</div>'; 
} 
} 
?>

为什么，当我做了var_dump($_POST['twitterq'])它始终是NULL？

UPDATE：

萤火虫给了我这个，不知道如何虽然修复：

enter image description here