Re: equivalent DataDiff function?

From: <wykoff_at_my-deja.com>
Date: Thu, 24 Jun 1999 18:11:29 GMT
Message-ID: <7ktsc5$d50$1@nnrp1.deja.com>

In article <37764d2a.66700410_at_newshost.us.oracle.com>, tkyte_at_us.oracle.com wrote:

> date arithmetic is so trivial that a specialized function like
datediff is not
> needed. Just subtract. You get the difference in days. Multiply by
24 --
> hours, multiply by 60 minutes, multiply by 60 -- seconds.
>

Wow, that was awesome. Thanks so much. Why, oh why can't I find this documented?

kim

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Share what you know. Learn what you don't. Received on Thu Jun 24 1999 - 13:11:29 CDT