xmlhttp.open();

Question

xmlhttp.open("POST","saveMneInfo.do?username="+username.value+"&ip="+SysInfo.ipstr+"&mac="+SysInfo.macaddrstr+"&cpu="+SysInfo.cpunumstr+"&hd="+hd,true);
这个open方法怎么用？

Answer 1

arg = "username=" + escape(username.value)+"&ip="+escape(SysInfo.ipstr)+"&mac="+escape(SysInfo.macaddrstr)+"&cpu="+escape(SysInfo.cpunumstr)+"&hd="+escape(hd);
xmlhttp.open("POST","saveMneInfo.do?" + escape(newe Date()),true);
xmlhttp.setRequestHeader("Content-Type", "application/x-www-form-urlencoded");  
xmlhttp.send(arg);

Answer 2

arg = "username=" + escape(username.value)+"&ip="+escape(SysInfo.ipstr)+"&mac="+escape(SysInfo.macaddrstr)+"&cpu="+escape(SysInfo.cpunumstr)+"&hd="+escape(hd);
xmlhttp.open("POST","saveMneInfo.do?" + escape(newe Date()),true);
xmlhttp.setRequestHeader("Content-Type", "application/x-www-form-urlencoded");  
xmlhttp.send(arg);

Answer 3

http://sophia-230.iteye.com/blog/174300
http://hi.baidu.com/gflovely/blog/item/f57556fb6cc7536d034f56ca.html

Answer 4

http://hi.baidu.com/gflovely/blog/item/f57556fb6cc7536d034f56ca.html