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oracle可以使用 lead、lag 函数来查询已有记录的下一条、上一条记录。
表结构如下:
如要查询Staffno是6-1102的前一条记录
select * from staff where staff_no=(select c.p from (select staff_no,lag(staff_no,1,0) over (order by staff_no) as p from staff) c where c.staff_no=‘6-1102‘)
结果:
STAFF_NO STAFF_NAME SEX
---------- -------------------- --- -
6-1076 梁柄聪 男
1 rows selected
如要查询其后一条记录
select * from staff where staff_no=(select c.n from (select staff_no,lead(staff_no,1,0) over (order by staff_no) as n from staff) c where c.staff_no=‘6-1102‘)
结果:
STAFF_NO STAFF_NAME SEX
---------- -------------------- --- -
6-1103 余志伟 男
1 rows selected
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原文地址:http://www.cnblogs.com/wjwen/p/5135187.html