Is the data in a partition compressed - oracle

This query will give me if a compression has been marked for compression
Select *
From All_Tab_Partitions
It came up in discussion that a Partition marked for compression can actually contain uncompressed data if it is loaded with FAST_LOAD.
I want to only issue compression commands for uncompressed partitions.
How can I find out if the data inside a partition is compressed or not?

Compression may be enabled on a partition but it is only used if data was created from a direct-path insert. Use sampling and the function
DBMS_COMPRESSION.GET_COMPRESSION_TYPE to estimate the percent of rows
compressed per partition.
Schema
create table compression_test
(
a number,
b varchar2(100)
) partition by range(a)
(
partition p_all_compressed values less than (2) compress,
partition p_none_compressed values less than (3) compress,
partition p_half_compressed values less than (4) compress
);
insert /*+ append */ into compression_test partition (p_all_compressed)
select 1, '0123456789' from dual connect by level <= 100000;
commit;
insert into compression_test partition (p_none_compressed)
select 2, '0123456789' from dual connect by level <= 100000;
commit;
insert /*+ append */ into compression_test partition (p_half_compressed)
select 3, '0123456789' from dual connect by level <= 50000;
commit;
insert into compression_test partition (p_half_compressed)
select 3, '0123456789' from dual connect by level <= 50000;
commit;
Estimate Code
--Find percent compressed from sampling each partition.
declare
v_percent_compressed number;
begin
--Loop through partitions.
for partitions in
(
select table_owner, table_name, partition_name
from dba_tab_partitions
--Enter owner and table_name here.
where table_owner = user
and table_name = 'COMPRESSION_TEST'
) loop
--Dynamic SQL to sample a partition and test for compression.
execute immediate '
select
round(sum(is_compressed)/count(is_compressed) * 100) percent_compressed
from
(
--Compression status of sampled rows.
--
--Numbers are based on constants from:
--docs.oracle.com/cd/E16655_01/appdev.121/e17602/d_compress.htm
--Assumption: Only basic compression is used.
--Assumption: Partitions are large enough for 0.1% sample size.
select
case dbms_compression.get_compression_type(
user,
'''||partitions.table_name||''',
rowid,
'''||partitions.partition_name||'''
)
when 4096 then 1
when 1 then 0
end is_compressed
from '||partitions.table_owner||'.'||partitions.table_name||'
partition ('||partitions.partition_name||') sample (0.1)
)
' into v_percent_compressed;
dbms_output.put_line(rpad(partitions.partition_name||':', 31, ' ')||
lpad(v_percent_compressed, 3, ' '));
end loop;
end;
/
Sample Output
P_ALL_COMPRESSED: 100
P_HALF_COMPRESSED: 55
P_NONE_COMPRESSED: 0

Related

Oracle counting rows in a PARTITION

I have the following setup, which includes PARTITIONS. Is there a query I can use that will provide a count for each PARTITION within the table..
I prefer not to have a possible estimate by gathering statistics as opposed to the actual count(*). Note the PARTITION name can be renamed!!
Below is my test CASE. Thanks to all who answer.
ALTER SESSION SET NLS_DATE_FORMAT = 'MMDDYYYY HH24:MI:SS';
CREATE TABLE dts (
dt DATE
)
PARTITION BY RANGE (dt)
INTERVAL (NUMTODSINTERVAL(7,'DAY'))
(
PARTITION OLD_DATA values LESS THAN (TO_DATE('2022-01-01','YYYY-MM-DD'))
);
INSERT into dts(dt)
select to_date (
'01-08-2022','mm-dd-yyyy' ) +
( level / 24 ) dt
from dual
connect by level <= ( 24 + ( 24 *
(to_date('01-15-2022' ,'mm-dd-yyyy') - to_date('01-08-2022','mm-dd-yyyy') )
)
) ;
SELECT table_name,
partition_name,
num_rows
FROM user_tab_partitions
WHERE table_name not like 'BIN$%'
ORDER BY table_name, partition_name;
TABLE_NAME PARTITION_NAME NUM_ROWS
DTS OLD_DATA -
DTS SYS_P415755 -
DTS SYS_P415756 -
Try this one:
declare
c integer;
begin
for aPart in (select partition_name FROM user_tab_partitions where table_name = 'DTS') loop
execute immediate 'select count(*) from DTS PARTITION ('||aPart.partition_name||')' INTO c;
DBMS_OUTPUT.PUT_LINE(aPart.partition_name || ' ' || c || ' rows');
end loop;
end;
select table_name ,Partition_name, to_number(extractvalue(xmltype(dbms_xmlgen.getxml('select /*+ parallel(a,8) */
count(*) c from '||table_name||' partition ('||partition_name||') a ')),'/ROWSET/ROW/C')) as count
from user_tab_partitions
TABLE_NAME PARTITION_NAME COUNT
DTS OLD_DATA 0
DTS SYS_P415799 167
DTS SYS_P415800 25
Oracle provides a handy function PMARKER exact for this purpose
SELECT DBMS_MVIEW.PMARKER(p.rowid) PMARKER, count(*) cnt, min(dt), max(dt)
from dts p
group by DBMS_MVIEW.PMARKER(p.rowid)
order by 1;
PMARKER CNT MIN(DT) MAX(DT)
---------- ---------- ------------------- -------------------
74312 167 08.01.2022 01:00:00 14.01.2022 23:00:00
74313 25 15.01.2022 00:00:00 16.01.2022 00:00:00
Note that you need not know the partition name, the partition key column value lets you access the partition using the partition extended names:
Example for the first partition
select count(*) from dts partition for (DATE'2022-01-08');
COUNT(*)
----------
167
You can rely on optimizer statistics for a perfect count, as long as you're using the default sample size and algorithm.
begin
dbms_stats.gather_table_stats
(
ownname => user,
tabname => 'DTS',
estimate_percent => dbms_stats.auto_sample_size
);
end;
/
If you run the above PL/SQL block, your original query against USER_TAB_PARTITIONS will return the correct NUM_ROWS. Since version 11g, Oracle scans the entire table to calculate statistics. While it uses an approximation for counting things like non-distinct values and histograms, it's trivial for the algorithm to get a completely accurate row count.
The manual is not super clear about this behavior, but you can put it together from the manual and other articles that discuss how the new algorithm works. From the "Gathering Optmizer Statistics" chapter of the "SQL Tuning Guide":
To maximize performance gains while achieving necessary statistical
accuracy, Oracle recommends that the ESTIMATE_PERCENT parameter use
the default setting of DBMS_STATS.AUTO_SAMPLE_SIZE. In this case,
Oracle Database chooses the sample size automatically. This setting
enables the use of the following:
A hash-based algorithm that is much faster than sampling
This algorithm reads all rows and produces statistics that are nearly
as accurate as statistics from a 100% sample. The statistics computed
using this technique are deterministic.
Most likely, you don't even need to specify the ESTIMATE_PERCENT => DBMS_STATS.AUTO_SAMPLE_SIZE argument. It is extremely unlikely for someone to set that preference for a table or the system. You can use the below query to see how your statistics are typically gathered. Most likely the query will return "DBMS_STATS.AUTO_SAMPLE_SIZE":
select dbms_stats.get_prefs(pname => 'ESTIMATE_PERCENT', ownname => user, tabname => 'DTS')
from dual;

Prevent Oracle indexes turn USABLE after INSERT

I have an interval partitioned table. I'd like to create a local index unusable by default to make all inserts work fast. I want manually turn indexes usable after insert. Currently every insert which creates new partition makes correspondend local index usable. How can I prevent it?
UPDATE: I'm working with Oracle 11.2
create table T (
part integer
,nm number
)
partition by range (part)
interval (1)
(partition empty values less than (1));
create index t_loc_idx on T (part, nm) local unusable
insert into T
select 1, rownum from user_tables
insert into T
select 2, rownum from user_tables
select index_name, partition_name, status from user_ind_partitions
where index_name = 'T_LOC_IDX'
What we can observe:
INDEX_NAME PARTITION_NAME STATUS
T_LOC_IDX EMPTY UNUSABLE
T_LOC_IDX SYS_P8744 USABLE
T_LOC_IDX SYS_P8745 USABLE
Keep interval partition local indexes unusable by creating a partial index and setting indexing off:
create table T (
part integer
,nm number
)
partition by range (part)
interval (1)
(partition empty values less than (1))
INDEXING OFF /* <-- New */;
create index t_loc_idx on T (part, nm) INDEXING PARTIAL /* <-- New */ local unusable ;
insert into T
select 1, rownum from user_tables where rownum <= 10;
insert into T
select 2, rownum from user_tables where rownum <= 10;
select index_name, partition_name, status from user_ind_partitions
where index_name = 'T_LOC_IDX';
New results:
INDEX_NAME PARTITION_NAME STATUS
========== ============== ======
T_LOC_IDX EMPTY UNUSABLE
T_LOC_IDX SYS_P643 UNUSABLE
T_LOC_IDX SYS_P644 UNUSABLE

Create partition in an indexed table

I have a table which holds data for 12 hours. Every 5 minutes, it keeps deleting data which is more than 12 hours old and adds new data. It has almost 15-20 million rows. I want to create partition by hour and also index the table on column(time_stamp), to make the row fetching faster.
I will obviously do interval or range partitioning, but found that interval partitioning doesn't work on indexed table. So please help me with the syntax so that oracle creates 12 partitions and automatically adds new one when new time_stamp data is added which is after first 12 hours. I have already got a procedure to delete oldest partition which i will use so that there is always 12 hours of data.
I am stating the columns below.
CustomerId,ApplicationId,Time_Stamp,Service
I have tried to come up with this, but don't know how it will create new partitions
CREATE TABLE local_table
(customerid VARCHAR2(30),
applicationid VARCHAR2(30),
time_stamp TIMESTAMP,
service VARCHAR2(30))
PARTITION BY RANGE(time_stamp)
(
PARTITION t1 VALUES LESS THAN(TO_TIMESTAMP('2015-02-25 00:00:00.0','YYYY-MM-DD HH24:MI:SS.ff')),
PARTITION t2 VALUES LESS THAN(TO_TIMESTAMP('2015-02-25 01:00:00.0','YYYY-MM-DD HH24:MI:SS.ff')),
PARTITION t3 VALUES LESS THAN(TO_TIMESTAMP('2015-02-25 02:00:00.0','YYYY-MM-DD HH24:MI:SS.ff')),
PARTITION t4 VALUES LESS THAN(TO_TIMESTAMP('2015-02-25 03:00:00.0','YYYY-MM- DD HH24:MI:SS.ff')),
PARTITION t5 VALUES LESS THAN(TO_TIMESTAMP('2015-02-25 04:00:00.0','YYYY-MM-DD HH24:MI:SS.ff')),
PARTITION t6 VALUES LESS THAN(TO_TIMESTAMP('2015-02-25 05:00:00.0','YYYY-MM-DD HH24:MI:SS.ff')),
PARTITION t7 VALUES LESS THAN(TO_TIMESTAMP('2015-02-25 06:00:00.0','YYYY-MM-DD HH24:MI:SS.ff')),
PARTITION t8 VALUES LESS THAN(TO_TIMESTAMP('2015-02-25 07:00:00.0','YYYY-MM-DD HH24:MI:SS.ff')),
PARTITION t9 VALUES LESS THAN(TO_TIMESTAMP('2015-02-25 08:00:00.0','YYYY-MM-DD HH24:MI:SS.ff')),
PARTITION t10 VALUES LESS THAN(TO_TIMESTAMP('2015-02-25 09:00:00.0','YYYY-MM-DD HH24:MI:SS.ff')),
PARTITION t11 VALUES LESS THAN(TO_TIMESTAMP('2015-02-25 10:00:00.0','YYYY-MM-DD HH24:MI:SS.ff')),
PARTITION t12 VALUES LESS THAN(TO_TIMESTAMP('2015-02-25 11:00:00.0','YYYY-MM-DD HH24:MI:SS.ff'))
);
create index index_time_stamp on local_table(TIME_STAMP);
I am using- Oracle Database 11g Enterprise Edition Release 11.2.0.4.0 - 64bit
Create table with autopartitiong and LOCAL (partitioning) index.
The local_partitioned_index clauses let you specify that the index is partitioned on the same columns, with the same number of partitions and the same partition bounds as table. Oracle Database automatically maintains local index partitioning as the underlying table is repartitioned.
CREATE TABLE local_table
(customerid VARCHAR2(30),
applicationid VARCHAR2(30),
time_stamp TIMESTAMP,
service VARCHAR2(30))
PARTITION BY RANGE(time_stamp)
INTERVAL(NUMTODSINTERVAL(1, 'HOUR'))
(PARTITION t1 VALUES LESS THAN(TO_TIMESTAMP('2015-02-25 00:00:00.0','YYYY-MM-DD HH24:MI:SS.ff'))
);
CREATE INDEX index_time_stamp on local_table(TIME_STAMP) LOCAL;
SELECT *
FROM user_tab_partitions;
INSERT INTO local_table VALUES('1', 'a', sysdate, 'b');
SELECT *
FROM user_tab_partitions;
INSERT INTO local_table VALUES('2', 'c', sysdate + 1/1440, 'd');
SELECT *
FROM user_tab_partitions;
INSERT INTO local_table VALUES('3', 'e', sysdate + 1/24, 'f');
SELECT *
FROM user_tab_partitions;
The INTERVAL clause of the CREATE TABLE statement establishes interval
partitioning for the table. You must specify at least one range
partition using the PARTITION clause. The range partitioning key value
determines the high value of the range partitions, which is called the
transition point, and the database automatically creates interval
partitions for data beyond that transition point. The lower boundary
of every interval partition is the non-inclusive upper boundary of the
previous range or interval partition.
For example, if you create an interval partitioned table with monthly
intervals and the transition point at January 1, 2010, then the lower
boundary for the January 2010 interval is January 1, 2010. The lower
boundary for the July 2010 interval is July 1, 2010, regardless of
whether the June 2010 partition was previously created. Note, however,
that using a date where the high or low bound of the partition would
be out of the range set for storage causes an error. For example,
TO_DATE('9999-12-01', 'YYYY-MM-DD') causes the high bound to be
10000-01-01, which would not be storable if 10000 is out of the legal
range.
Some quick DRAFT for your second question about DROP PARTITION. Examine and debug before uncomment ALTER TABLE. You can create scheduler job for run this block of code every hour.
DECLARE
l_pt_cnt NUMBER;
l_pt_name VARCHAR2(100);
l_minrowid ROWID;
l_mindate TIMESTAMP;
BEGIN
-- get partition count
SELECT count(*)
INTO l_pt_cnt
FROM user_tab_partitions
WHERE table_name = 'LOCAL_TABLE';
IF l_pt_cnt > 12 THEN
SELECT min(time_stamp)
INTO l_mindate
FROM LOCAL_TABLE;
-- get ROWID with min date
SELECT min(rowid)
INTO l_minrowid
FROM LOCAL_TABLE
WHERE time_stamp = l_mindate;
-- get name of partition with row with min date
SELECT subobject_name
INTO l_pt_name
FROM LOCAL_TABLE
JOIN user_objects
ON dbms_rowid.rowid_object(LOCAL_TABLE.rowid) = user_objects.object_id
WHERE LOCAL_TABLE.rowid = l_minrowid;
DBMS_OUTPUT.put_line('ALTER TABLE LOCAL_TABLE DROP PARTITION ' || l_pt_name );
--EXECUTE IMMEDIATE 'ALTER TABLE LOCAL_TABLE DROP PARTITION ' || l_pt_name;
END IF;
END;

Pipeline table function

I would like to understand why pipelined function is not returning any results
Any ideas what i am doing wrong here.
Giving credit where due to the Original Author
CREATE OR REPLACE PACKAGE MANAGE_SPACE AS
--Author Tanmay
g_tblspce_threshold number := 80;
TYPE tblespaces_record IS RECORD(
tablespace_name VARCHAR2(30),
percentage_used NUMBER
);
TYPE tblespaces_table IS TABLE OF tblespaces_record;
function list_tblspcs_excd_thresld
return tblespaces_table PIPELINED ;
END MANAGE_SPACE;
Package body
CREATE OR REPLACE PACKAGE BODY MANAGE_SPACE
AS
--Author Tanmay
FUNCTION list_tblspcs_excd_thresld
RETURN tblespaces_table PIPELINED
AS
tblspaces tblespaces_record;
BEGIN
for x in (SELECT a.tablespace_name tablespace_name,
ROUND (((c.BYTES - NVL (b.BYTES, 0)) / c.BYTES) * 100) percentage_used
into tblspaces
FROM dba_tablespaces a,
(SELECT tablespace_name,
SUM (BYTES) BYTES
FROM dba_free_space
GROUP BY tablespace_name
) b,
(SELECT COUNT (1) DATAFILES,
SUM (BYTES) BYTES,
tablespace_name
FROM dba_data_files
GROUP BY tablespace_name
) c
WHERE b.tablespace_name(+) = a.tablespace_name
AND c.tablespace_name(+) = a.tablespace_name
AND ROUND (((c.BYTES - NVL (b.BYTES, 0)) / c.BYTES) * 100) > g_tblspce_threshold
ORDER BY NVL (((c.BYTES - NVL (b.BYTES, 0)) / c.BYTES), 0) DESC)
loop
PIPE ROW (tblspaces);
end loop;
return;
END list_tblspcs_excd_thresld;
END MANAGE_SPACE;
executing this package does not return any rows
SQL> select * from table(MANAGE_SPACE.list_tblspcs_excd_thresld())
2
SQL> /
TABLESPACE_NAME PERCENTAGE_USED
------------------------------ ---------------
What am i doing wrong
You need to populate an array. The easiest way to do this is to use the BULK COLLECT syntax. Then loop round the array and pipe out rows.
Here is my revised version of your package.
CREATE OR REPLACE PACKAGE BODY MANAGE_SPACE
AS
FUNCTION list_tblspcs_excd_thresld
RETURN tblespaces_table PIPELINED
AS
-- collection type not record type
tblspaces tblespaces_table;
BEGIN
select *
-- populate the collection
bulk collect into tblspaces
from
(SELECT a.tablespace_name tablespace_name,
ROUND (((c.BYTES - NVL (b.BYTES, 0)) / c.BYTES) * 100) percentage_used
FROM dba_tablespaces a,
(SELECT tablespace_name,
SUM (BYTES) BYTES
FROM dba_free_space
GROUP BY tablespace_name
) b,
(SELECT COUNT (1) DATAFILES,
SUM (BYTES) BYTES,
tablespace_name
FROM dba_data_files
GROUP BY tablespace_name
) c
WHERE b.tablespace_name(+) = a.tablespace_name
AND c.tablespace_name(+) = a.tablespace_name
AND ROUND (((c.BYTES - NVL (b.BYTES, 0)) / c.BYTES) * 100) > g_tblspce_threshold
ORDER BY NVL (((c.BYTES - NVL (b.BYTES, 0)) / c.BYTES), 0) DESC);
-- loop round the collection
for i in 1..tblspaces.count()
loop
PIPE ROW (tblspaces(i));
end loop;
return;
END list_tblspcs_excd_thresld;
END MANAGE_SPACE;
Here is the output:
SQL> select * from table(MANAGE_SPACE.list_tblspcs_excd_thresld())
2 /
TABLESPACE_NAME PERCENTAGE_USED
------------------------------ ---------------
SYSTEM 100
SYSAUX 91
APEX_2614203650434107 90
EXAMPLE 87
USERS 82
SQL>
#a_horse_with_no_name makes a good point regarding memory consumption. Collections are held in the session's memory space, the PGA. In this particular case it is unlikely that a database would have enough tablespaces to blow the PGA limit. But for other queries which might have much larger result sets there is the BULK COLLECT ... LIMIT syntax, which allows us to fetch manageable chunks of the result set into our collection. Find out more.

Oracle: how to drop a subpartition of a specific partition

I am using an oracle 11 table with interval partitioning and list subpartitioning like this (simplified):
CREATE TABLE LOG
(
ID NUMBER(15, 0) NOT NULL PRIMARY KEY
, MSG_TIME DATE NOT NULL
, MSG_NR VARCHAR2(16 BYTE)
) PARTITION BY RANGE (MSG_TIME) INTERVAL (NUMTOYMINTERVAL (1,'MONTH'))
SUBPARTITION BY LIST (MSG_NR)
SUBPARTITION TEMPLATE (
SUBPARTITION login VALUES ('FOO')
, SUBPARTITION others VALUES (DEFAULT)
)
(PARTITION oldvalues VALUES LESS THAN (TO_DATE('01-01-2010','DD-MM-YYYY')));
How do I drop a specific subpartitition for a specific month without knowing the (system generated) name of the subpartition? There is a syntax "alter table ... drop subpartition for (subpartition_key_value , ...)" but I don't see a way to specify the month for which I am deleting the subpartition. The partition administration guide does not give any examples, either. 8-}
You can use the metadata tables to get the specific subpartition name:
SQL> insert into log values (1, sysdate, 'FOO');
1 row(s) inserted.
SQL> SELECT p.partition_name, s.subpartition_name, p.high_value, s.high_value
2 FROM user_tab_partitions p
3 JOIN
4 user_tab_subpartitions s
5 ON s.table_name = p.table_name
6 AND s.partition_name = p.partition_name
7 AND p.table_name = 'LOG';
PARTITION_NAME SUBPARTITION_NAME HIGH_VALUE HIGH_VALUE
--------------- ------------------ ------------ ----------
OLDVALUES OLDVALUES_OTHERS 2010-01-01 DEFAULT
OLDVALUES OLDVALUES_LOGIN 2010-01-01 'FOO'
SYS_P469754 SYS_SUBP469753 2012-10-01 DEFAULT
SYS_P469754 SYS_SUBP469752 2012-10-01 'FOO'
SQL> alter table log drop subpartition SYS_SUBP469752;
Table altered.
If you want to drop a partition dynamically, it can be tricky to find it with the ALL_TAB_SUBPARTITIONS view because the HIGH_VALUE column may not be simple to query. In that case you could use DBMS_ROWID to find the subpartition object_id of a given row:
SQL> insert into log values (4, sysdate, 'FOO');
1 row(s) inserted.
SQL> DECLARE
2 l_rowid_in ROWID;
3 l_rowid_type NUMBER;
4 l_object_number NUMBER;
5 l_relative_fno NUMBER;
6 l_block_number NUMBER;
7 l_row_number NUMBER;
8 BEGIN
9 SELECT rowid INTO l_rowid_in FROM log WHERE id = 4;
10 dbms_rowid.rowid_info(rowid_in =>l_rowid_in ,
11 rowid_type =>l_rowid_type ,
12 object_number =>l_object_number,
13 relative_fno =>l_relative_fno ,
14 block_number =>l_block_number ,
15 row_number =>l_row_number );
16 dbms_output.put_line('object_number ='||l_object_number);
17 END;
18 /
object_number =15838049
SQL> select object_name, subobject_name, object_type
2 from all_objects where object_id = '15838049';
OBJECT_NAME SUBOBJECT_NAME OBJECT_TYPE
--------------- --------------- ------------------
LOG SYS_SUBP469757 TABLE SUBPARTITION
As it turns out, the "subpartition for" syntax does indeed work, though that seems to be a secret Oracle does not want to tell you about. :-)
ALTER TABLE TB_LOG_MESSAGE DROP SUBPARTITION FOR
(TO_DATE('01.02.2010','DD.MM.YYYY'), 'FOO')
This deletes the subpartition that would contain MSG_TIME 2010/02/01 and MSG_NR FOO. (It is not necessary that there is an actual row with this exact MSG_TIME and MSG_NR. It throws an error if there is no such subpartition, though.)
Thanks for the post - it was very useful for me.
One observation though on the above script to identify the partition and delete it:
The object_id returned by dbms_rowid.rowid_info is not the object_id of the all_objects table. It is actually the data_object_id. It is observed that usually these ids match. However, after truncating the partitioned table several times, these ids diverged in my database. Hence it might be reasonable to instead use the data_object_id to find out the name of the partition:
select object_name, subobject_name, object_type
from all_objects where data_object_id = '15838049';
From the table description of ALL_OBJECTS:
OBJECT_ID Object number of the object
DATA_OBJECT_ID Object number of the segment which contains the object
http://docs.oracle.com/cd/B19306_01/appdev.102/b14258/d_rowid.htm
In the sample code provided in the above link, DBMS_ROWID.ROWID_OBJECT(row_id) is used instead to derive the same information that is given by dbms_rowid.rowid_info. However, the documentation around this sample mentions that it is a data object number from the ROWID.
Examples
This example returns the ROWID for a row in the EMP table, extracts
the data object number from the ROWID, using the ROWID_OBJECT function
in the DBMS_ROWID package, then displays the object number:
DECLARE object_no INTEGER; row_id ROWID; ... BEGIN
SELECT ROWID INTO row_id FROM emp
WHERE empno = 7499; object_no := DBMS_ROWID.ROWID_OBJECT(row_id); DBMS_OUTPUT.PUT_LINE('The obj. # is
'|| object_no); ...

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