假设这两个列分别称为DT和PRICE，并且假设只有一个“东西”的价格是您监视的（否则您需要一个GROUP BY子句）：

select min(dt) as dt, min(price) keep (dense_rank first order by dt) as price
from   your_table
where  price > ( select min(price) keep (dense_rank first order by dt)
                 from   your_table
               )

https://docs.oracle.com/cd/B19306_01/server.102/b14200/functions056.htm

使用MATCH_RECOGNIZE子句的解决方案（需要Oracle 12及更高版本）。
我在WITH子句中创建了测试数据。这不是解决方案的一部分; SQL查询在WITH子句之后的SELECT TICKER, ....处开始
PATTERN中的不情愿匹配中的问号会触发JDBC驱动程序，因此无法从SQL Developer运行此查询;它需要在SQL*Plus或类似的前端中运行。（解决方法是将b*?更改为b*，并在DEFINE子句中添加：b as b.price <= a.price .）
为了更好地说明MATCH_RECOGNIZE的灵活性，我假设可能有几个“ticker”，每个都有其起始日期（最早的价格日期），查询将查找每个ticker第一次出现的价格高于原始价格的情况。

with
     test_data ( ticker, dt, price ) as (
       select 'XYZ', to_date('20170322', 'yyyymmdd'), 109.89 from dual union all
       select 'XYZ', to_date('20170321', 'yyyymmdd'), 107.02 from dual union all
       select 'XYZ', to_date('20170320', 'yyyymmdd'), 109.25 from dual union all
       select 'XYZ', to_date('20170317', 'yyyymmdd'), 108.44 from dual union all
       select 'XYZ', to_date('20170316', 'yyyymmdd'), 108.53 from dual union all
       select 'XYZ', to_date('20170315', 'yyyymmdd'), 107.94 from dual union all
       select 'XYZ', to_date('20170314', 'yyyymmdd'), 106.83 from dual union all
       select 'XYZ', to_date('20170313', 'yyyymmdd'), 110.02 from dual union all
       select 'XYZ', to_date('20170310', 'yyyymmdd'), 107.31 from dual union all
       select 'XYZ', to_date('20170309', 'yyyymmdd'), 107.54 from dual union all
       select 'XYZ', to_date('20170308', 'yyyymmdd'), 107.67 from dual union all
       select 'XYZ', to_date('20170307', 'yyyymmdd'), 108.98 from dual
     )
select ticker, dt, price
from   test_data
match_recognize (
  partition by ticker
  order     by dt
  measures  c.dt as dt, c.price as price
  one row per match
  pattern   ( ^ a b*? c )
  define    c as c.price > a.price
)
;

TICKER  DT            PRICE
------  ----------  -------
XYZ     2017-03-13   110.02

1 row selected.