處理方法其實很簡單、只要: alter user <username> identified by <same password>; 這個操作後,恢複正常了 下面作個簡單測試:SQL> conn hr/hrERROR:ORA-28002: the password will expire within 2 daysConnected.SQL> conn / as sysdbaConnected.SQL> alter
1、Where子句中的串連順序: ORACLE採用自下而上的順序解析WHERE子句。 根據這個原理,表之間的串連必須寫在其他WHERE條件之前, 那些可以過濾掉最大數量記錄的條件必須寫在WHERE子句的末尾。 舉例: (低效)select … from table1 t1 where t1.sal > 300 and t1.jobtype = '0001' and 20 < (select count(*) from table1 t2 where t2.
--分頁select * from(select t1.ename ,rownum rn from (select * from emp) t1 where rownum<=10) where rn>=6; --開發一個包create or replace package tespackage astype test_cursor is ref cursor;end tespackage;--分頁預存程序create or replace procedure
今天寫oracle語句時發現一個奇怪的問題,糾結了很久沒有找到原因。求大牛解惑。問題精簡如下:select seq,info ,topic from system.help where seq in (select distinct seq from system.help where seq=20); -----(1)與 select seq,info ,topic from system.help where seq
4.1 編寫DEBUG 程式包(常式)CREATE OR REPLACE PACKAGE yang_debug ASPROCEDURE out(p_comments IN VARCHAR2, p_varible IN VARCHAR2);PROCEDURE Erase;END yang_debug;CREATE OR REPLACE PACKAGE BODY yang_debug ASPROCEDURE out(p_comments IN VARCHAR2, p_varible IN
Oracle複雜查詢 資料分組,分組函數max, min, avg, sum, count查詢工資最高的人的姓名 select ename from emp where sal = (select max(sal) from emp);給所有低於平均工資的員工薪水上漲10% update emp set sal = (select avt(sal) from emp)*10% where sal<(select avt(sal) from emp);grout by 和 having