oracle 甲骨文儒略日
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Oracle Julian day of year
提问by Prmejc
how can I select Julian day of year in Oracle database?
如何在 Oracle 数据库中选择儒略日?
I tried: select to_char(sysdate, 'J') from dual; Which gives me the number of days since January 1, 4712 BC. But I would need the number of days since 1.1. of current year.
我试过: select to_char(sysdate, 'J') from dual; 这给了我自公元前 4712 年 1 月 1 日以来的天数。但我需要自 1.1 以来的天数。当年。
回答by álvaro González
If you check the TO_CHAR (datetime)documentation you get a link to "Format Models"with a comprehensive list of available formats. I guess you want this:
如果您检查TO_CHAR(日期时间)文档,您将获得一个指向“格式模型”的链接,其中包含可用格式的完整列表。我猜你想要这个:
DDD
Day of year (1-366)
DDD
一年中的第几天 (1-366)
回答by Christian
One way would be to use:
一种方法是使用:
select sysdate - trunc(sysdate,'yyyy') from dual
'Trunc' cuts everything except the year and returns 01/01/2014, subtracted by the sysdate returns numbers of days since 1st of january.
'Trnc' 删除除年份以外的所有内容并返回 01/01/2014,减去 sysdate 返回自 1 月 1 日以来的天数。
回答by Jeff Mergler
SELECT TO_CHAR(SYSDATE, 'DDD') from DUAL;
回答by acboone
Use sql select trunc(sysdate)+1 - trunc(sysdate,'yyyy') from dual
. you will get an even number
使用sql select trunc(sysdate)+1 - trunc(sysdate,'yyyy') from dual
. 你会得到一个偶数