Oracle数据库如何获取多条结果集中的第一条或某一条

--获取多条结果集中的第一条，通常会排下序之类的，随你方便，如果你不排序直接用括号里的就行
select * from 
	(select * from table 
	 where 1 = 1 
	 order by value1)
where rownum = 1;

select * from table where rownum = 1 order by value1;

select * from 
	( select rownum id,t1.* from
		 (select t2.* from t2 order by t2.某列) 
	t1)
where id=第几条;

where id=第几条

where id = 5

SELECT *
  FROM (
       SELECT ROW_NUMBER() OVER(
       PARTITION BY  ZA40011_3.URIAGE_DENPYO_NO,ZA40011_3.URIAGE_DENPYO_ROW_NO  
       ORDER BY  ZA40011_3.KAKUTEI_SURYO_CS,
                  ZA40011_3.KAKUTEI_SURYO_BL,
                  ZA40011_3.KAKUTEI_SURYO_BT,
                  ZA40011_3.SYUKKA_SHIJI_NO,
                  ZA40011_3.SYUKKA_SHIJI_ROW_NO  DESC) LEV,
                ZA40011_3 .*
       FROM  WT_001_ZA40011_22070_2 ZA40011_2,
             WT_001_ZA40011_22070_3 ZA40011_3,
             JT_GYOMU_URIAGE_DETAIL URIAGE_DETAIL
       WHERE 
       ZA40011_2.KAISYA_CODE = URIAGE_DETAIL.KAISYA_CODE
           AND ZA40011_2.URIAGE_DENPYO_NO = URIAGE_DETAIL.URIAGE_DENPYO_NO
           AND ZA40011_2.URIAGE_DENPYO_ROW_NO = URIAGE_DETAIL.ROW_NO
           AND URIAGE_DETAIL.KAISYA_CODE = ZA40011_3.KAISYA_CODE
           AND URIAGE_DETAIL.URIAGE_DENPYO_NO = ZA40011_3.URIAGE_DENPYO_NO
           AND URIAGE_DETAIL.ROW_NO = ZA40011_3.URIAGE_DENPYO_ROW_NO
           AND ZA40011_3.KEPPIN_RIYU_CODE IS NOT NULL
        )
 WHERE LEV = 1
-- ORDER BY KAKUTEI_SURYO_CS DESC