我刚刚用这个脚本建立了一个网站:
<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.3.2/jquery.min.js"></script>
<script>
function loadpage(page)
{
document.getElementById("pageContent").innerHTML="Yükleniyor...";
if (window.XMLHttpRequest)
{// code for IE7+, Firefox, Chrome, Opera, Safari
xmlhttp=new XMLHttpRequest();
}
else
{// code for IE6, IE5
xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange=function()
{
if (xmlhttp.readyState==4 && xmlhttp.status==200)
{
document.getElementById("pageContent").innerHTML=xmlhttp.responseText;
}
}
xmlhttp.open("GET",page,true);
xmlhttp.send();
}
</script>
Ajax可以加载任何页面,但是仍然有一个问题:当加载包含HTML表单的页面时,一旦单击“Submit”就会离开主页,我的意思是我不能通过Ajax发送表单变量,只能通过使用“href”和前面提到的loadpage()函数传递表单变量。
如何获取表单输入的值并发送到另一个PHP文件?
As you are using jQuery, you can do:
$('form').submit(funciton() {
var data = $(this).serialize();
// Call Ajax
return false;
});
I advice you to read about: http://api.jquery.com/category/ajax/ and http://api.jquery.com/serialize/.
you can use jQuery.
$(document).ready(function(){
$("#div_load").load("page.html");
});
whit this code you can open any page (Ex: page.html) in any div(Ex:div whit id=div_load).
and for sending data use it:
$(".class_div").click(function(){
$.post("ajax.php",
{
name:"naser",
age:"23"
},
function(data,status){
// do something when done
});
});