FAQ
no, it should get the correct size allocated to all the partitions

select sum(bytes)/(1024*1024*1024) from dba_segments where segment_name =;

SUM(BYTES)/(1024*1024*1024)

18

select partition_name, sum(bytes)/(1024*1024*1024) from dba_segments where
segment_name = group by partition_name;

PARTITION_NAME SUM(BYTES)/(1024*1024*1024)
------------------------------ ---------------------------
TAB_2002_1Q 2.375
TAB_2002_2Q 2.25
TAB_2002_3Q 2
TAB_2002_4Q 2.25
TAB_2003_1Q 2.625
TAB_2003_2Q 2.125
TAB_2003_3Q 2
TAB_2003_4Q .125
TAB_HIST 2.125
TAB_MAX1 .125

Babu
----- Original Message -----
To: "Multiple recipients of list ORACLE-L"
Sent: Wednesday, October 15, 2003 3:04 PM
Well, if the table is partitioned you'll get a nice fat zero....
On 10/15/2003 03:49:36 PM, Stephen Lee wrote:

How about:
select sum(bytes) from dba_segments where segment_name =
'TABLE_NAME';
-----Original Message-----

Hi All,
Could somebody help me in finding the actual size of an oracle
table in
GB.

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