当前位置:编程学习 > 网站相关 >>

使用ajax调用webservice

注意,使用ajax调用webservice时,尽量使用ie浏览器,如果使用chrome或者是firefox浏览器,很可能会出现以下异常


[java]
2013-6-1 11:10:02 com.sun.xml.internal.ws.transport.http.server.WSHttpHandler handleExchange 
警告: Cannot handle HTTP method: OPTIONS 

2013-6-1 11:10:02 com.sun.xml.internal.ws.transport.http.server.WSHttpHandler handleExchange
警告: Cannot handle HTTP method: OPTIONS
1、服务器端代码的书写(可以参考使用jdk调用webservice中的代码,两者是基本相同的)

2、ajax_webservice.html


[html] 
<html> 
    <head> 
        <title>通过ajax调用WebService服务</title> 
        <script> 
             function getXhr(){ 
                var xhr = null; 
                if(window.XMLHttpRequest){ 
                    //非ie浏览器 
                    xhr = new XMLHttpRequest(); 
                }else{ 
                    //ie浏览器 
                    xhr = new ActiveXObject('Microsoft.XMLHttp'); 
                } 
                return xhr; 
            } 
          var xhr =getXhr(); 
            function sendMsg(){ 
                var name = document.getElementById('name').value; 
                //服务的地址 
                var wsUrl = 'http://127.0.0.1:6790/hello'; 
                 
                //请求体 
                 var soap= '<soapenv:Envelope xmlns:soapenv="http://schemas.xmlsoap.org/soap/envelope/" xmlns:q0="http://webservice.njupt.com/" xmlns:xsd="http://www.w3.org/2001/XMLSchema" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance">' 
             +'<soapenv:Body><q0:sayHello><arg0>'+name+'</arg0> </q0:sayHello></soapenv:Body></soapenv:Envelope>'; 
                          
                //打开连接 
                xhr.open('POST',wsUrl,true); 
                 
                //重新设置请求头 
                xhr.setRequestHeader("Content-Type","text/xml;charset=UTF-8"); 
                 
                //设置回调函数 
                xhr.onreadystatechange = _back; 
                 
                //发送请求 
                xhr.send(soap); 
            } 
             
            function _back(){ 
                if(xhr.readyState == 4){ 
                    if(xhr.status == 200){ 
                            //alert('调用Webservice成功了'); 
                            var ret = xhr.responseXML; 
                            var msg = ret.getElementsByTagName('return')[0]; 
                            document.getElementById('showInfo').innerHTML = msg.text; 
                            //alert(msg.text); 
                        } 
                } 
            } 
        </script> 
    </head> 
    <body> 
            <input type="button" value="发送SOAP请求" onclick="sendMsg();"> 
            <input type="text" id="name"> 
            <div id="showInfo"> 
            </div> 
    </body> 
</html> 

<html>
 <head>
  <t

补充:Web开发 , 其他 ,
CopyRight © 2022 站长资源库 编程知识问答 zzzyk.com All Rights Reserved
部分文章来自网络,