页面返回时保持状态的2种解决思路

1.在session里面放一个map集合,里面存放各种条件,封装存取方法.感觉这种方法很稳定,但是session里面放太多东西不太好.

2.将页面地址及所有参数解析出来,拼装成新地址.那么新地址就可以保存所有的参数了.但是这会有些小问题,比如说转发,那么获得的新地址就会是最初的那个,而不是最终转发的那个.解决方法:重定向.

相关代码:

　HttpServletRequest r = ServletActionContext.getRequest();

　StringBuffer query = new StringBuffer("");
   String uri = r.getRequestURI();

   ActionContext c = ActionContext.getContext();
   Map params = c.getParameters();

   for (Iterator iter = params.keySet().iterator(); iter.hasNext();) {
    Object paramKey = iter.next();
    if (paramKey == null) {
     continue;
    }
    Object paramValue = params.get(paramKey);

    if (paramValue instanceof Object[]) {
     query = query.append(extractToStringBuffer(paramKey, paramValue));
    } else {
     query.append("&");
     query.append(paramKey.toString());
     query.append("=");
     query.append(paramValue.toString());
    }
   }

   uri = uri + query;

private String extractToStringBuffer(Object param, Object values)
   throws UnsupportedEncodingException {
  String sb = "";

  Object[] objs = (Object[]) values;
  if (objs == null)
   return "";
  if (objs.length == 0)
   return "";

  for (int i = 0; i < objs.length; i++) {
   sb = sb + "&";
   sb = sb + param.toString();
   sb = sb + "=";
   Object v = objs[i];

   sb = sb + v.toString();

  }

  return sb;
 }