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oracle的substr函数的用法[@more@]取得字符串中指定起始位置和长度的字符串 substr( string, start_position, [ length ] )如:substr('This is a test', 6, 2) would return 'is'substr('This is a test', 6) would return 'is a test'substr('TechOnTheNet', -3, 3) would return 'Net'substr('TechOnTheNet', -6, 3) would return 'The'

avl,lag函数巧用[@more@]有如下的資料:C_DATE NO20070501 120070501 220070501 320070501 420070501 620070501 820070501 920070502 120070502 2=====================我現在想要得到如下的結果:C_DATE Min_NO Max_NO20070501 1 420070501 6 620070501 8 920070502 1 2=====================就是說　如果NO不連續了就換行。請問有好的方法碼？？select c_date, nvl(lag(e) over(partition by c_date order by s), minn) min_no, nvl(S, maxn) max_no from (select c_date, lag(no, 1) over(partition by c_dat

rownum问题讨论[@more@]rownum问题讨论 我在进行分页查询测试时遇到如下问题：我用了两种方式测试第一种：用between and 的方式：--测试1：执行0.741sselect * from(select rownum rn,A.object_id,A.object_name from all_objects A ) where rn between 15 and 35 --测试2：执行0.772sselect * from(select rownum rn,A.object_id,A.object_name from all_objects A ) where rn between 25000 and 25020 --测试3：执行0.781sselect * from(select rownum rn,A.object_id,A.object_name from all_objects A ) where rn between 46000 and 4