Home > Backend Development > PHP Tutorial > 在ajax中用POST方法提交的XML串,服务器端怎么回显到客户端

在ajax中用POST方法提交的XML串,服务器端怎么回显到客户端

WBOY
Release: 2016-06-13 13:53:31
Original
894 people have browsed it

在ajax中用POST方法提交的XML串,服务器端如何回显到客户端?
符上原代码:
nbsp; HTML   PUBLIC   "-//W3C//DTD   HTML   4.0   Transitional//EN ">

 


      New   Document  
   
   
   
   
 
<script> <br /> var xmlhttp; <br /> function createXmlRequest(){ <br /> if(window.ActiveXObject){ <br /> xmlhttp=new window.ActiveXObject( "Microsoft.XMLHTTP "); <br /> }else{ <br /> if(window.XMLHttpRequest){ <br /> xmlhttp=new XMLHttpRequest(); <br /> }else{ <br /> alert( "error "); <br /> } <br /> } <br /> var url= "postxml.php? "+new Date().getTime(); <br /> xmlhttp.open( "POST ",url,true); <br /> xmlhttp.setRequestHeader( "Content-Type ", "application/x-www-form-urlencoded; "); <br /> xmlhttp.onreadystatechange=show; <br /> xmlhttp.send(createXml()); <br /> } <br /> function createXml(){ <br /> var xml= " <pets> "; <br /> var options=document.getElementById( "petTypes ").childNodes; <br /> var option=null; <br /> for(var i=0;i <options.length;i++){ <br /> option=options[i]; <br /> if(option.selected){ <br /> xml=xml+ " <type> "+option.value+ " "; <br /> } <br /> } <br /> alert(xml+ " "); <br /> return xml+ " "; <br /> <br /> } <br /> function show(){ <br /> if(xmlhttp.readyState==4){ <br /> if(xmlhttp.status==200){ <br /> var response=document.getElementById( "response "); <br /> if(response.hasChildNodes()){ <br /> response.removeChild(response.childNodes[0]); <br /> } <br /> document.getElementById( "response ").appendChild(document.createTextNode(xmlhttp.responseText)); <br /> } <br /> } <br /> } <br /> </script>
 
   

select   OPtion


   

   
   
   

    Server   response
   
test

 



------解决方案--------------------
客户端:下面这个函数里面应该添加接收后的处理.
function show(){
if(xmlhttp.readyState==4){
if(xmlhttp.status==200){
var return_value = xmlhttp.responseXML;
//然后你再对返回的这个值进行处理.
}
}
服务器端:postxml.php中
根据你所获取的参数进行输出一个XML格式的东西.


我也是刚接触.如有不对,敬请谅解.
Related labels:
source:php.cn
Statement of this Website
The content of this article is voluntarily contributed by netizens, and the copyright belongs to the original author. This site does not assume corresponding legal responsibility. If you find any content suspected of plagiarism or infringement, please contact admin@php.cn
Popular Tutorials
More>
Latest Downloads
More>
Web Effects
Website Source Code
Website Materials
Front End Template