Question

如何将输出限制为仅获得ORACLE中的第一行。我尝试过FETCH first 1 rows only;，但它却给sql无法正确结束的错误消息

当前输出

 70 19-APR-18   Base Line Date
 71 20-JUN-19   Target Date
 73 23-JUN-18

查询

SELECT  EXEC_TRACKER_SESSION_SEQ,
(CASE WHEN tsm.SESSION_SEQ IS NULL THEN tsm.TAGETDATE ELSE s.SESSION_DATE END) as SESSION_DATE
,tsm.NOTES 
FROM TRACKER_SESSION_MAP tsm
LEFT JOIN session s ON tsm.session_seq = s.session_seq
WHERE tsm.TRACKER_SEQ =244
order by TRACKER_SESSION_SEQ ASC

必需的输出

70  19-APR-18   Base Line Date

Answer 1

您需要

WHERE ROWNUM=1

Oracle没有top X函数，但它始终知道rownumber是什么，因此请充分利用。

Answer 2

将rownum=1用作：

select * from
( 
SELECT  EXEC_TRACKER_SESSION_SEQ,
(CASE WHEN tsm.SESSION_SEQ IS NULL THEN tsm.TAGETDATE ELSE s.SESSION_DATE END) as SESSION_DATE
,tsm.NOTES 
FROM TRACKER_SESSION_MAP tsm
LEFT JOIN session s ON tsm.session_seq = s.session_seq
WHERE tsm.TRACKER_SEQ =244
order by TRACKER_SESSION_SEQ ASC
) 
where rownum=1;

或者，您可以将row_number()函数用作：

select * from
(
 SELECT  EXEC_TRACKER_SESSION_SEQ,
(CASE WHEN tsm.SESSION_SEQ IS NULL THEN tsm.TAGETDATE ELSE s.SESSION_DATE END) as SESSION_DATE
,tsm.NOTES, row_number() over (order by day) as rn 
FROM TRACKER_SESSION_MAP tsm
LEFT JOIN session s ON tsm.session_seq = s.session_seq
WHERE tsm.TRACKER_SEQ =244
order by TRACKER_SESSION_SEQ ASC
) 
where rn=1;

只要您使用fetch..rows，

Oracle 12c语句就可以使用。

如何在Oracle中查找将行限制为前1个

2 个答案: