" select max(su.rwjhjssj) rwjhjssj,su.sid from ( select sub.sid sid,sub.mid mid , sub.rwjhjssj rwjhjssj from prp_jdgl_jdglwbslcbjh " + " jh right join prp_jdgl_jdglwbslcbjh_sub sub on sub.mid=jh.sid where jh.tid in ( select sid from prp_htgl_zcbhtxx where orgid in (" + orgids + ")" + " ) and sub.rwjhjssj < '" + nowdate + "'" + " GROUP BY sub.mid,rwjhjssj desc ) su group by su.mid ";
這里有兩個思路：
1、先按照要求分組並按照時間從大到小排序；這里數據不會實現真正分組而是按照分組的數據放在一起並實現時間的降序排序；例如：GROUP BY sub.mid,rwjhjssj desc
2、最后在進行分組在利用聚合函數取出時間最大的或者滿足要求的數據；例如：group by su.mid
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