求排名 mysql构建rownum字段

create table t(s number);insert into t values(90);insert into t values(50);insert into t values(20);commit;问题：给出一个数，求出这个数在这个表中的排名假如：给出100，返回1给出90，返回1给出80，返回2给出40，返回3给出10，返回4
提示：下文中的40是给出的数1   oracle数据库下select nvl((    select rn    from    (        select fir.*,rownum as rn from        (select t.*,40 as value from t order by s desc) fir    ) sec    where value-s=    (        select min(value-s) as cut        from        (            select fir.*,rownum as rn from            (select t.*,40 as value from t order by s desc) fir        )sec         where value-s>=0    ) ),--参数一(select count(*)+1  from t)    --参数二)  as  rank from dual
2   mysql数据库下mysql中没有oracle的rownum，所以需要自己构建一个相似于oracle的rownum字段select ifnull((select rownumfrom( select *,40 value from (select  a.*,(select count(*)from t where t.s>=a.s) as rownum from t a) fir     --构建类似的rownum字段) secwhere value-s=( select min(value-s) as cut from ( select *,40 value from (select  a.*,(select count(*)from t where t.s>=a.s) as rownum from t a) fir   --构建类似的rownum字段 ) sec  where value-s>=0) ),(select count(*)+1 as s from t)) as rank