将Unixtime转换为Datetime SQL(Oracle) [英] Convert Unixtime to Datetime SQL (Oracle)

问题描述

我有一个datetime字段(P_DT)，我想返回所有结果，其中P_DT大于输入的unix时间戳.

I have a datetime field(P_DT) and I would like to return all results where P_DT is greater then an input unix timestamp.

Oracle有任何内置功能可以提供帮助吗?

Does Oracle have any built in functions that can help?

在搜索中，我找到了DateTime到Unix的结果，但没有Unix到DateTime的结果...

In my searchs I find resuts for DateTime to Unix but no Unix to DateTime...

推荐答案

没有内置函数.但是写一个相对容易.由于Unix时间戳是自1970年1月1日以来的秒数

There are no built-in functions. But it's relatively easy to write one. Since a Unix timestamp is the number of seconds since January 1, 1970

CREATE OR REPLACE FUNCTION unix_ts_to_date( p_unix_ts IN NUMBER )
  RETURN DATE
IS
  l_date DATE;
BEGIN
  l_date := date '1970-01-01' + p_unix_ts/60/60/24;
  RETURN l_date;
END;

您可以看到它被称为

SQL> select unix_ts_to_date( 1336822620 ) from dual;

UNIX_TS_TO_DATE(133
-------------------
2012-05-12 11:37:00

这篇关于将Unixtime转换为Datetime SQL(Oracle)的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！