幾個月前，因為項目需求，我寫了下面的三個AJax相關的函數。發布出來和大家分享。
第一個是用來無刷新加載一段Html
第二個是把表單數據轉換成一串請求字符串
第三個是結合函數一和函數二的無刷新提交表單實現。
還有一點要提到的是，無刷新表單提交，還不能對文件上傳進行處理，這個主要是因為浏覽器的安全設置。目前無刷新的上傳，一般是用iframe來實現的。關於這個，我們在google裡搜索能找到很多。
網上雖然已經有很多優秀的AJax的類和函數了，但是或許我這幾個函數對大家還有點用處，於是我就發布出來了。
可以在這裡下載。
[復制此代碼]CODE:
//@desc load a page(some Html) via XMLhttp,and display on a container
//@param url the url of the page will load,such as "index.PHP"
//@param request request string to be sent,such as "action=1&name=surfchen"
//@param method POST or GET
//@param container the container object,the loaded page will display in container.innerHtml
//@usage
// AJaxLoadPage('index.PHP','action=1&name=surfchen','POST',document.getElementById('my_home'))
// suppose there is a Html element of "my_home" id,such as "<span id='my_home'></span>"
//@author SurfChen <surfchen@gmail.com>
//@url http://www.surfchen.org/
//@license http://www.gnu.org/licenses/gpl.Html GPL
function AJaxLoadPage(url,request,method,container)
{
method=method.toUpperCase();
var loading_msg='Loading...';//the text shows on the container on loading.
var loader=new XMLHttpRequest;//require Cross-Browser XMLHttpRequest
if (method=='GET')
{
urls=url.split("?");
if (urls[1]=='' || typeof urls[1]=='undefined')
{
url=urls[0]+"?"+request;
}
else
{
url=urls[0]+"?"+urls[1]+"&"+request;
}
request=null;//for GET method,loader should send NULL
}
loader.open(method,url,true);
if (method=="POST")
{
loader.setRequestHeader("Content-Type","application/x-www-form-urlencoded");
}
loader.onreadystatechange=function(){
if (loader.readyState==1)
{
container.innerHtml=loading_msg;
}
if (loader.readyState==4)
{
container.innerHtml=loader.responseText;
}
}
loader.send(request);
}
//@desc transform the elements of a form object and their values into request string( such as "action=1&name=surfchen")
//@param form_obj the form object
//@usage formToRequestString(document.form1)
//@notice this function can not be used to upload a file.if there is a file input element,the func will take it as a text input.
// as I know,because of the security,in most of the browsers,we can not upload a file via XMLhttp.
// a solution is iframe.
//@author SurfChen <surfchen@gmail.com>
//@url http://www.surfchen.org/
//@license http://www.gnu.org/licenses/gpl.Html GPL
function formToRequestString(form_obj)
{
var query_string='';
var and='';
//alert(form_obj.length);
for (i=0;i<form_obj.length ;i++ )
{
e=form_obj[i];
if (e.name!='')
{
if (e.type=='select-one')
{
element_value=e.options[e.selectedIndex].value;
}
else if (e.type=='checkbox' || e.type=='radio')
{
if (e.checked==false)
{
break;
}
element_value=e.value;
}
else
{
element_value=e.value;
}
query_string+=and+e.name+'='+element_value.replace(/\&/g,"%26");
and="&"
}
}
return query_string;
}
//@desc no refresh submit(ajax) by using AJaxLoadPage and formToRequestString
//@param form_obj the form object
//@param container the container object,the loaded page will display in container.innerHtml
//@usage AJaxFormSubmit(document.form1,document.getElementById('my_home'))
//@author SurfChen <surfchen@gmail.com>
//@url http://www.surfchen.org/
//@license http://www.gnu.org/licenses/gpl.Html GPL
function AJaxFormSubmit(form_obj,container)
{
AJaxLoadPage(form_obj.getAttributeNode("action").value,formToRequestString(form_obj),form_obj.method,container)
}