Size of Rows

Hi,
How can we calculate the size of rows or row. I mean suppose i have a table with 50 million rows (lets assume the size is 20 GB). Now out of this how can i calculate the size of say 22million rows. The 22 million rows may be selected by firing a query that satisfy some criteria.
Kindly help me.
Thanks in advance.

Assuming you're selecting a reasonable and representative fraction of the rows, I'd suspect that you'd just want something like
data volume = (# rows you're selecting / # rows in table) * size of table
If you use the CBO and your statistics are reasonably accurate, you could also
SELECT <<num rows you're selecting>> * avg_row_len
FROM user_tables
WHERE table_name = <<table name>>Justin

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Hi,
I am observing some row size anomaly which is causing a table to grow enormously. The table is a partitioned table. The size of row as calculated from dba_tab_columns is 2338 bytes. The average length of the row is 379 bytes, calculated by querying dba_tables. The table is analyzed just before calculating any of the values.
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Thanks in advance for your valuable time.....
Database: 10.2.0.4
OS: UNIX
Environment: DWH, Production
Regards,
Joy

Hi,
I am having a similar question.
The Verison is
BANNER
Oracle Database 11g Enterprise Edition Release 11.1.0.7.0 - 64bit Production
PL/SQL Release 11.1.0.7.0 - Production
CORE 11.1.0.7.0 Production
TNS for Solaris: Version 11.1.0.7.0 - Production
NLSRTL Version 11.1.0.7.0 - Production
The Blocksize is
SQL> SELECT name,value
FROM v$parameter
WHERE name = 'db_block_size'
NAME VALUE
db_block_size 8192
1 row selected.
The lenght in total of the table is
SQL> select sum(data_length) sum_data_length from dba_tab_columns
where owner = 'SIEBEL' and table_name = 'EIM_ACCNT_UT'
SUM_DATA_LENGTH
27131
1 row selected.
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where owner = 'SIEBEL' and table_name = 'EIM_ACCNT_UT'
NUM_ROWS AVG_ROW_LEN CHAIN_CNT
173 356 5223
1 row selected.
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1.) is the value of 27131 Bytes as records lenght as I counted it, rigth or wrong ?
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Thanks Pravan for your reply. But Its still not so usefull for me. Can you please give some clear idea about what you wanna say ??
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select * from ........................................
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