I have a table in SQL Server that needs its normal pk index to be replaced by a clustered index on two fields. These other fields are not part of the primary key.
I need to do something similar in Oracle. As far as I know, this can be done using index-ordered tables, but my guess is these indexes are only constructed on the primary key.
Is there any way I can get a similar behavior to SQLServer's clustered index in Oracle?
Index organized tables are the concept of oracle which is near to clustered indexes in sql server. I found a discussion about the topic in oracle forum and one on asktom.
My question is: why do you want to adapt the behavior? What is the benefit you whant to obtain?
A clustered index in sql server is mostly the primary key index. Rowdata is stored into the index node. The conecpt on oracle to store row data into the index is an index organized table.
On oracle the iot is used to avoid a second lookup for row data into the table after index lookup.
The purpose of clustered index on sql server is to store the rowdata. A table can have only one clustered index. This index will hold the rowdata. Any other index is a non clustered index.
IMHO the concept of clustered index is bound to sql server data storage and there is no need to rebuild this behavior in oracle. Oracle has other concepts to store data.
Answer:
A regular index on oracle is all to solve your problem.
Related
I am coming from MS sql server where clustered and non clustered index used to play important role.
But looks like there are nothing of this sort under oracle. I am sure there must be some equivalent of
clustered and non clustered index in oracle . Can somebody throw light on this ?
When we say create index in oracle is it equivalent to non clustered index ?
Index Organised table in Oracle stores data of the whole table sorted on the basis of say - primary key. So, this is the closest thing to have as clustered Index in Oracle, other than that all other indexes are non-clustered but the Index key for all other indexes are sorted too with Rowid in-front of them which points to the actual data.
Is there a big performance hit when running SQL on a partitioned Oracle table where the SQL does not reference the column that is used for partitioning?
It depends. If your WHERE clause is not based on any index column your query will be very slow, because Oracle has to do a Full-Table-Scan.
If you have a globally defined index on columns in WHERE clause, it does not matter whether you access only one partition or all.
But you get a degradation when your table has many partions and locally defined index which are used by WHERE clause. Assume your table has 50 partitions, and locally defined index, then a query which does not specify the partition (either by WHERE condition or explicitly by partition name) has to scan 50 individual indexes likewise.
When i create a primary key in oracle table, why does it create a 'clustered' index by default. What is the reason for this automatic creation of a clustered index on creation of a primary key? is it just the Oracle designer's preference that he designed oracle in this way?
Oracle will create an index to police an unique constraint where no pre-existing index is suitable. Without the index, Oracle would need to serialize operations (such as a table lock) whenever someone tries to insert or delete a row (or update the PK).
Contrarily to MS-SQL Server, this index is not clustered on heap tables (default table organization), i.e. this index won't change the underlying table structure and natural order. The rows won't be reordered when Oracle creates the index. The index will be a B-tree index and will exist as a separate entity where each entry points to a row in the main table.
Oracle doesn't have clustered index as MS SQL, however indexed-organized tables share some properties with cluster-indexed tables. The PK is an integral part of such tables and has to be specified during creation.
(Oracle also has table clusters, but they are a completely different concept).
Creating Index is basic functionality of Primary key, it is also in SQL Server and MySQL, Clustered Index makes your searches faster.
The Database Engine automatically creates a unique index to enforce the uniqueness of the PRIMARY KEY constraint. If a clustered index does not already exist on the table or a nonclustered index is not explicitly specified, a unique, clustered index is created to enforce the PRIMARY KEY constraint.
Read this:
http://www.sqlskills.com/blogs/kimberly/the-clustered-index-debate-continues/
I'm currenly developing on Oracle. I have several tables for which I defined FOREIGN KEY constraints. I have already read this SQL Server-oriented and this MySQL-oriented questions but I could find none about Oracle.
So the question is always the same: in order to optimize query performance, for those columns for which I create a FOREIGN KEY constraint, do I also have to create an explicit secondary index? Doesn't Oracle automatically create an index on FOREIGN KEYed columns to boost performances during JOINs?
I usually perform queries in which the WHERE clause compare against those columns.
No, Oracle doesn't automatically create indexes on foreign key columns, even though in 99% of cases you probably should. Apart from helping with queries, the index also improves the performance of delete statements on the parent table.
How can I determine if an Oracle index is clustered or unclustered?
I've done
select FIELD from TABLE where rownum <100
where FIELD is the field on which is built the index. I have ordered tuples, but the result is wrong because the index is unclustered.
By default all indexes in Oracle are unclustered. The only clustered indexes in Oracle are the Index-Organized tables (IOT) primary key indexes.
You can determine if a table is an IOT by looking at the IOT_TYPE column in the ALL_TABLES view (its primary key could be determined by querying the ALL_CONSTRAINTS and ALL_CONS_COLUMNS views).
Here are some reasons why your query might return ordered rows:
Your table is index-organized and FIELD is the leading part of its primary key.
Your table is heap-organized but the rows are by chance ordered by FIELD, this happens sometimes on an incrementing identity column.
Case 2 will return sorted rows only by chance. The order of the inserts is not guaranteed, furthermore Oracle is free to reuse old blocks if some happen to have available space in the future, disrupting the fragile ordering.
Case 1 will most of the time return ordered rows, however you shouldn't rely on it since the order of the rows returned depends upon the algorithm of the access path which may change in the future (or if you change DB parameter, especially parallelism).
In both case if you want ordered rows you should supply an ORDER BY clause:
SELECT field
FROM (SELECT field
FROM TABLE
ORDER BY field)
WHERE rownum <= 100;
There is no concept of a "clustered index" in Oracle as in SQL Server and Sybase. There is an Index-Organized Table, which is similar but not the same.
"Clustered" indices, as implemented in Sybase, MS SQL Server and possibly others, where rows are physically stored in the order of the indexed column(s) don't exist as such in Oracle. "Cluster" has a different meaning in Oracle, relating, I believe, to the way blocks and tables are organized.
Oracle does have "Index Organized Tables", which are physically equivalent, but they're used much less frequently because the query optimizer works differently.
The closest I can get to an answer to the identification question is to try something like this:
SELECT IOT_TYPE FROM user_tables
WHERE table_name = '<your table name>'
My 10g instance reports IOT or null accordingly.
Index Organized Tables have to be organized on the primary key. Where the primary key is a sequence generated value this is often useless or even counter-productive (because simultaneous inserts get into conflict for the same block).
Single table clusters can be used to group data with the same column value in the same database block(s). But they are not ordered.