求一个Oracle函数 - Oracle / 开发
实现功能
输入"2008-12-21 12:20:20"
输出"2008-12-21 12:00:00"
我写的有问题
CREATE FUNCTION getdatehour(IDE date)
RETURN date as OED date;
BEGIN
OED:=cast(to_char(IDE,'YY')+'-'+to_char(IDE,'MM')+'-'+to_char(IDE,'DD')+' '+to_char(IDE,'HH24')+':00:00' as date);
return ODE;
END;
谁能给我个正确的函数啊
CREATE FUNCTION getdatehour(IDE date)
RETURN date as OED date;
BEGIN
OED:=substr(IDE,1,14)||00:00;
return ODE;
END;
CREATE OR REPLACE FUNCTION getdatehour(IDE date)
RETURN date
is
OED date;
BEGIN
OED:=trunc(IDE,'hh');
return OED;
END;
select getdatehour(sysdate) from dual;
1楼报错:
alter FUNCTION getdatehour(IDE date)
*
ERROR 1 :
ORA-00922:
SP2-0552: "00"
SQL code:
CREATE or replace FUNCTION getdatehour(IDE date)
RETURN date as
ODE date;
BEGIN
ODE := to_date(to_char(ide, 'yyyy-mm-dd hh24') || '00:00','yyyy-mm-dd hh24:mi:ss');
return ODE;
END;
测试代码:
SQL code:
declare
d date := sysdate;
begin
d := getdatehour(d);
db
相关问答:
我在开发的这个应用并发性比较高,oracle更新数据时候如何处理并发处理呢?各位大侠发表好的观点啊,在线等待!
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引用
oracle ......
谢谢唐人老大的指点,谢谢各位高手的指点!
安装10g Enterprise。上网下了MLDN的Oracle视频,跟着视频上的配置走,安装成功了,但是没完整地配置。
点击进入:Database Configuration Assistant
1. 创建数据库;
......
SELECT * from (SELECT A.*, ROWNUM RN from (SELECT * from t_out_logon) A WHERE ROWNUM <= 30)WHERE RN >= 21
SELECT A.* from (SELECT t.*,rownum rn from t_out_logon t WHERE ROWNUM <= 30) A WHERE ......