Apress.Expert.Oracle.Database.Architecture.9i.and.10g.Programming.Techniques.and.Solutions.Sep.2005

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CHAPTER 13 ■ PARTITIONING
Hash Partition Using Powers of Two
I mentioned earlier that the number of partitions should be a power of two. This is easily
observed to be true. To demonstrate, we’ll set up a stored procedure to automate the creation
of a hash partitioned table with N partitions (N will be a parameter). This procedure will construct
a dynamic query to retrieve the counts of rows by partition and then display the counts
and a simple histogram of the counts by partition. Lastly, it will open this query and let us see
the results. This procedure starts with the hash table creation. We will use a table named T:
ops$tkyte@ORA10G> create or replace
2 procedure hash_proc
3 ( p_nhash in number,
4 p_cursor out sys_refcursor )
5 authid current_user
6 as
7 l_text long;
8 l_template long :=
9 'select $POS$ oc, ''p$POS$'' pname, count(*) cnt ' ||
10 'from t partition ( $PNAME$ ) union all ';
11 begin
12 begin
13 execute immediate 'drop table t';
14 exception when others
15 then null;
16 end;
17
18 execute immediate '
19 CREATE TABLE t ( id )
20 partition by hash(id)
21 partitions ' || p_nhash || '
22 as
23 select rownum
24 from all_objects';
Next, we will dynamically construct a query to retrieve the count of rows by partition. It
does this using the “template” query defined earlier. For each partition, we’ll gather the count
using the partition-extended table name and union all of the counts together:
25
26 for x in ( select partition_name pname,
27 PARTITION_POSITION pos
28 from user_tab_partitions
29 where table_name = 'T'
30 order by partition_position )
31 loop
32 l_text := l_text ||
33 replace(
34 replace(l_template,