请问：getRequestDispatcher如何把参数也发送给下一个页面？
具体是这样的，当程序捕获到异常时，就转跳，同时要带错误信息tipMsg。我现在的代码是这样的，但是在businessAnalysis.jsp里用String   tipMsg   =   request.getParameter( "tipMsg ");得到的却是null，不知道咋回事。
代码如下:
try{
tmp   =   startDate.split( "- ");
st   =   tmp[0]+ "/ "+tmp[1]+ "/ "+tmp[2]   +   "   "   +   startHour   + ": "+   startMin   + ": "+   "00 ";
tmp   =   endDate.split( "- ");
et   =   tmp[0]+ "/ "+tmp[1]+ "/ "+tmp[2]   +   "   "   +   endHour   + ": "+   endMin   + ": "+   "00 ";
}catch(Exception   e){
tipMsg   =   "Error   Date   Format! ";
//System.exit(1);   will   close   the   server!!!
request.setAttribute( "searchDay ",startDate);
request.setAttribute( "tipMsg ",tipMsg);
//response.setAttribute( "tipMsg ",tipMsg);
request.getRequestDispatcher( "businessAnalysis.jsp ").forward(request,   response);  
//response.sendRedirect( "businessAnalysis.jsp?searchDay= "+startDate+ "&tipMsg= "+tipMsg);
}
关于这种动态的（根据判定条件决定如何转跳），带变量的页面转发，还有啥更好的办法么？为啥上面的代码不能把tipMsg传过去呢？
谢谢！

------解决方案--------------------
把
String tipMsg = request.getParameter( "tipMsg ");
改成
String tipMsg = request.getAttribute( "tipMsg ");
试试
------解决方案--------------------
request.getAttribute( "name ");
要不改成request.getRequestDispatcher( "businessAnalysis.jsp?tipmsg= "+tipmsg).forward(request, response);