Ajax从一个链接中获取多个值

我有一个链接，我想从中获取多个值，以便通过Ajax提交。那么，我该如何从该链接中选择streamitem_creator/target/content和type_id以传递到share.php中呢？

echo"<a title='Share ".$poster_name['fullusersname']."s status' href='include/share.php?streamitem_creator=".$streamitem_data['streamitem_creator']."&streamitem_target=".$_SESSION['id']."&streamitem_content=".$streamitem_data['streamitem_content']."&streamitem_type_id=4'/>Share</a>";

我已经提出了我的独奏用户数据-在链接

链接

 echo'<a class="sharelink" title="Share '.$poster_name['fullusersname'].'s status" href="#"
     data-streamitem_creator='.$streamitem_data['streamitem_creator'].'
     data-streamitem_target='.$_SESSION['id'].'
     data-streamitem_content='.$streamitem_data['streamitem_content'].'
     data-streamitem_type_id=4>Share</a>';

AJAX

$(document).ready(function(){
$('.sharelink').click(function(e){
var streamitem_creator = $(this).data('streamitem_creator');
var streamitem_target = $(this).data('streamitem_target');
var streamitem_content = $(this).data('streamitem_content');
var streamitem_type_id = $(this).data('streamitem_type_id');
$.ajax({
type: "GET",
url: "../include/share.php",
data: { streamitem_creator: streamitem_creator, streamitem_target: streamitem_target, streamitem_content: streamitem_content, streamitem_type_id: streamitem_type_id }, 
success: function(msg){
$('#result').html(msg);
e.preventDefault(); 
}
});
});
});