请问jquery ajax结合php这样提交数据防刷新对吗?如果不对请指出正确方法并加以说明谢谢!
</script>
window.onload = function () {
ajax_change_cont(1, '#ajax_cont');
}
function ajax_change_cont(type, cid) {
$(cid).html("<img src='/images/loading.gif' />");
$.ajax({
type: "POST",
url: "ajax/ajax_ind.aspx?type=" + escape(type),
success: function (msg) {
//msg是异步返回的数据
if (msg == "") {
if (type == 1) {
$(cid).html("异步返回的数据1")
} else {
$(cid).html("异步返回的数据2")
}
} else {
$(cid).html(msg);
}
}
});
}
</script>
<form>
<div id='ajax_cont'></div> //此处是异步返回数据要填充的地方
</form>