springmvc筛选发房间状态

Question

*
这要怎么筛选
mapper里写

<select id="showFindBystates" resultType="com.five.hotel.entity.Room" parameterType="java.lang.String">
    select rid as rId,roomtype.rtname as rTName,rprice as rPrice,rstates as rStates from room
     inner join roomtype on room.rtype=roomtype.rtid where 1=1
     <if test="_parameter!=null and r_parameter=''">
        and rstates=#{_parameter}
     </if>
</select>
controller里这样写
/**
 * 筛选房间
 * @param rStates
 * @param model
 * @return
 */
@RequestMapping("/search")
public String findRoom(Model model,HttpServletRequest request){
    String rStates=request.getParameter("rStates");
    List<Room> room=roomService.showFindBystates(rStates);
    model.addAttribute("Room",room);
    return "room/room";
}

但是不知道要怎么筛选
一直报错

Answer 1

and rstates=#{_parameter}

错了换成 r_parameter ！=‘’