Ajax通过post方法向Servlet提交信息

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Js代码
    @H_404_6@ <scripttype="text/javascript">
  1. varreq;
  2. varcontent="id=123&name=gavin&address=WorshintonDC.USA";
  3. functionvalidate(){
  4. //创建一个XMLHttpRequest对象req
  5. if(window.XMLHttpRequest){
  6. //IE7,Firefox,Opera支持
  7. req=newXMLHttpRequest();
  8. }elseif(window.ActiveXObject){
  9. //IE5,IE6支持
  10. newActiveXObject("Microsoft.XMLHTTP");
  11. }
  12. req.open("post",url,true);
  13. req.setRequestHeader("Content-Type","application/x-www-form-urlencoded");
  14. req.onreadystatechange=callback;
  15. //send函数发送请求,参数
  16. req.send(content);
  17. </script>

在servlet中:

Java代码
    response.setContentType("text/html");
  1. response.setHeader("Cache-Control","no-store");
  2. response.setHeader("Pragma","no-cache");
  3. response.setDateHeader("Expires",0);
  4. Stringput=null;
  5. for(Enumeration<String>e=request.getParameterNames();e.hasMoreElements();){
  6. Stringh=(String)e.nextElement();
  7. Stringc=(String)request.getParameter(h);
  8. put+=c;
  9. out.write("messageis:"+put);

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