如何只刷新成功的div而不是使用Ajax刷新整个页面?
$('a.add-to-cart').click(function (e) {
var url = "<?php echo createItemLink($row['id']);?>";
e.preventDefault();
$.ajax({
type: 'post',
url: "<?php echo SITEURL;?>/ajax/addtoCart.php",
data: 'itemid=<?php echo $row['id'];?>&price=<?php echo $row['price'];?>&name=<?php echo $row['title'];?>',
success: function (msg) {
$("#status").fadeIn(300).delay(800).fadeOut(300).ajaxComplete(function (event, request, settings) {
$(this).html(msg);
$(location).attr('href',url);
});
}
});
});
答案 0 :(得分:2)
让我们假设div id是theDiv
success: function (msg) {
$("#status").fadeIn(300).delay(800).fadeOut(300).ajaxComplete(function (event, request, settings) {
$('#theDiv').html(msg);
});
}
答案 1 :(得分:1)
这里div类是要用msg <
替换的div的类名 success: function (msg) {
<br> $(.divclass).html(msg);
<br> });
<br>}