SELECT table_name as Table_Name, row_cnt as Row_Count, SUM(mb) as Size_MB
FROM
(SELECT in_tbl.table_name, to_number(extractvalue(xmltype(dbms_xmlgen.getxml('select count(*) c from ' ||ut.table_name)),'/ROWSET/ROW/C')) AS row_cnt , mb
FROM
(SELECT CASE WHEN lob_tables IS NULL THEN table_name WHEN lob_tables IS NOT NULL THEN lob_tables END AS table_name , mb
FROM (SELECT ul.table_name AS lob_tables, us.segment_name AS table_name , us.bytes/1024/1024 MB FROM user_segments us
LEFT JOIN user_lobs ul ON us.segment_name = ul.segment_name ) ) in_tbl INNER JOIN user_tables ut ON in_tbl.table_name = ut.table_name ) GROUP BY table_name, row_cnt ORDER BY 3 DESC;``
Above query will give, Table_name, Row_count, Size_in_MB(includes lob column size) of specific user.
If you just want to calculate the schema size without tablespace free space and indexes :
select
sum(bytes)/1024/1024 as size_in_mega,
segment_type
from
dba_segments
where
owner='<schema's owner>'
group by
segment_type;
For all schemas
select
sum(bytes)/1024/1024 as size_in_mega, owner
from
dba_segments
group by
owner;
SELECT DS.TABLESPACE_NAME, SEGMENT_NAME, ROUND(SUM(DS.BYTES) / (1024 * 1024)) AS MB
FROM DBA_SEGMENTS DS
WHERE SEGMENT_NAME IN (SELECT TABLE_NAME FROM DBA_TABLES) AND SEGMENT_NAME='YOUR_TABLE_NAME'
GROUP BY DS.TABLESPACE_NAME, SEGMENT_NAME;
You probably want
SELECT sum(bytes)
FROM dba_segments
WHERE owner = <<owner of schema>>
If you are logged in as the schema owner, you can also
SELECT SUM(bytes)
FROM user_segments
That will give you the space allocated to the objects owned by the user in whatever tablespaces they are in. There may be empty space allocated to the tables that is counted as allocated by these queries.
select T.TABLE_NAME, T.TABLESPACE_NAME, t.avg_row_len*t.num_rows from dba_tables t
order by T.TABLE_NAME asc
See e.g. http://www.dba-oracle.com/t_script_oracle_table_size.htm for more options
Source: Stackoverflow.com