RE: avg_row_len
- From: "Powell, Mark" <mark.powell2@xxxxxx>
- To: "oracle-l@xxxxxxxxxxxxx" <oracle-l@xxxxxxxxxxxxx>
- Date: Fri, 9 Mar 2012 17:48:16 +0000
There is also overhead that will be required to hold the new rows so
avg_row_len X num_rows X overhead_factor will provide a better estimate. I
suggest estimating the new table size for the full load and then look to see
how full the current table is. Some of your data may fit into the existing
space. Finally you have to take a look at your net space need in relation to
the table extent allocation method in use.
-----Original Message-----
From: oracle-l-bounce@xxxxxxxxxxxxx [mailto:oracle-l-bounce@xxxxxxxxxxxxx] On
Behalf Of Timo Raitalaakso
Sent: Thursday, March 08, 2012 4:08 PM
To: oracle-l@xxxxxxxxxxxxx
Subject: Re: avg_row_len
The estimate would be something like your formula. Check also LAST_ANALYZED and
does NUM_ROWS match count(*).
http://docs.oracle.com/cd/E11882_01/server.112/e25513/statviews_2117.htm#i1592091
|AVG_ROW_LEN*| |NUMBER| Average length of a row in the table
(in bytes)
How full are the used column data types populated? Does_ALL_TAB_COLUMNS
AVG_COL_LEN match DATA_LENGTH
_http://docs.oracle.com/cd/E11882_01/server.112/e25513/statviews_2103.htm#I1020277
Will the future 600k row be poplated with similar data? You will not know. But
it is an estimate.
--
Timo Raitalaakso
http://rafudb.blogspot.com
8.3.2012 22:43, Zelli, Brian kirjoitti:
> So I can see that the average row length in my table is 263. Is that bytes?
> And if I want to estimate another 600,000 rows I multiply the 600,000 by 263?
>
> ciao,
> Brian
>
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