使用xmlhttp.send()方法的参数来传递大容量的数据

    这个标题不知道怎么取才好,其实要实现的功能就是在客户端要将大量的数据传递到服务端,我们知道url的长度是有限制的,如果使用url的参数来传递数据,只能传递很少的关键信息,但有时候我们要传递大量的信息,比如一张单据中的所有商品的商品编号等等.
    Ajax技术中的xmlhttp.send()方法提供我们传送大量数据的功能,要使用该功能,首先要将要传到服务端的数据封装在一个xml对象中,在将这个xml对象的xml属性做为send()方法的参数就可以.
    客户端代码:

                    var guidArray=guidlist.split(",");
                    
var xmlDom=new ActiveXObject("MSXML2.DOMDocument");
                    xmlDom.loadXML(
"");
                    
var domRoot= xmlDom.createElement("NewDataSet");
                    xmlDom.appendChild(domRoot);
                    
for(var i=0;i<guidArray.length;i++)
                    {
                       
var node= xmlDom.createElement("guid");
                       node.text
=guidArray[i];
                       domRoot.appendChild(node);                       
                    }
                    
                    
var xmlhttp=new ActiveXObject("MSXML2.XMLHTTP");                    
                    xmlhttp.onreadystatechange
=function(){
                            
if (xmlhttp.readyState==4)
                            {
                              
if (xmlhttp.status==200)
                              {
                                 alert(
"处理完成");
                              }
                              
else
                              {
                                 alert(xmlhttp.responseText);  
                              }
                            }
                                  
                    }
                    xmlhttp.Open(
"POST","fmSmsPostProcess.aspx?sStatus="+escape(sStatus)+"&sProcessResult="+escape(document.getElementById("btProcessResult").value)+
                                         
"&sReStore="+escape(document.getElementById("btReStore").value),true);
                    xmlhttp.setRequestHeader(
"Content-Type","text/xml");                                         
                    xmlhttp.send(xmlDom.xml);
服务端需要从请求流中,解析出客户端传入的xml的内容,服务端代码:
                System.IO.Stream instream = Page.Request.InputStream;
                BinaryReader br 
= new BinaryReader(instream,System.Text.Encoding.UTF8);
                
byte[] byt = br.ReadBytes((int)instream.Length);
                
string sXml = System.Text.Encoding.UTF8.GetString(byt);

                System.Xml.XmlDocument xmlDoc 
= new System.Xml.XmlDocument();
                xmlDoc.LoadXml(sXml);

                XmlElement xe
= xmlDoc.DocumentElement;
                
for(int i=0;i< xe.ChildNodes.Count;i++)
                {
                    sGuid 
= sGuid + ",'" + xe.ChildNodes[i].InnerText+"'";
                }

posted on 2006-03-27 15:33  DoNet鸟  阅读(4395)  评论(3编辑  收藏  举报

导航