Re: How much space do my data occupy?
Date: Mon, 20 Feb 1995 21:58:07
Message-ID: <odysscci.175.0015F89C_at_teleport.com>
In article <5gD5ou-1xEB_at_becker.ping.de> udo_at_becker.ping.de (Udo Becker) writes:
>Path: news.teleport.com!usenet.reed.edu!news.reed.edu!usenet.ee.pdx.edu!cs.uoregon.edu!news.uoregon.edu!vixen.cso.uiuc.edu!howland.reston.ans.net!Germany.EU.net!ping.de!becker.ping.de!udo
>Date: 20 Feb 1995 22:13:00 +0200
>From: udo_at_becker.ping.de (Udo Becker)
>Newsgroups: comp.databases.oracle
>Message-ID: <5gD5ou-1xEB_at_becker.ping.de>
>References: <3grp5p$c4k_at_lucy.infi.net> <3h7amc$s7q_at_newsbf02.news.aol.com>
> <outputD482pu.A7B_at_netcom.com>
>Subject: How much space do my data occupy?
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>Hi,
>how can I determine how much space my data occupy?
>From COLS I can compute how many bytes at maximum I need to represent one
>row of a given table T:
> select sum(data_length) from cols where table_name = "T"; (LEN)
>(Our Oracle says it needs 22 bytes to represent an integer value. I can't
>imagine this is true.)
>I could multiply this by the total amount of rows in my table (AM), giving
>LEN*AM.
>Furthermore I can determine how many extents are realy needed for my table
>and how many bytes they occupy:
> select sum(bytes) from dba_extents where segment_name = "T";
>In most cases the result is less than LEN*AM. This is even true if my
>table only consists of fixed length columns, e.g. integers.
>By selecting from DBA_FREE_SPACE I can compute how many bytes are free in
>my tablespaces.
>But I haven't found a way to tell how many bytes are free, resp. occupied,
>in my extents.
>Udo Becker
>--
> __
>| | |__) Dipl.-Inf. Udo Becker phone: ++49-2335-62562 (priv.)
>|__| |__) Haus Hove 17 ++49-2335-92-7974 (off.)
> D-58300 Wetter fax: ++49-2335-92-7811 (off.)
>## CrossPoint v3.0 ##
I believe that Integers do not always take 22 bytes. Certainly, if the
integer was large enough it could. Since numbers are stored in base 100 with
an offset I think that the following would give you the maximum size for your
numbers:
y = ((log10( your largest number) + number of didgets of precision)/2)+1
for integers the precision would be 0.
I think that this will give you a better estimate. Jim Kennedy Received on Mon Feb 20 1995 - 21:58:07 CET