Is an index clustered or unclustered in Oracle? - oracle

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.

Related

Is there any use to create index on all the table columns in oracle?

In our one of production database, we have 4 column table and there are no PK,UK constraints on it. only one notnull constraint on one column. The inserts are slow on this table and when I checked the indexes , there is one index which is built on all columns.
It is a normal table and not IOT. I really don't see a need of all column index, but wondering why the developers has created it?
Appreciate your thoughts?
It might be usefull, i.e. if you (mainly) query all columns oracle doesn't have to access the table at all, but can get all the data from the index. Though inserts take longer because a larger index has to be maintained by the dbms everytime.
One case where it could be useful is,
Say for example, you are trying to check the existence of records in this table and for that you have to have joins on all four columns. So in such a case if you have written a correlated query like below,
SELECT <something>
FROM table_1 t1
WHERE EXISTS
(SELECT 1 FROM table_t2 t2 where t1.c1=t2.c1 and t1.c2=t2.c2 and t1.c3=t2.c3 and t1.c4=t2.c4)
Apart from above case, it looks an error to me from developer's side.
Indexes are good to better query optimization but causes slow updates/inserts because the indexes needs to be updated at each modification.
If these tables first use is querying and inserts happens only in a specific periods like a batch at the beginning or the end of the day only, then you can remove the indexes before updating tables and then restore them.
In addition, all the queries all these tables need to be analysed to see which indexes are useful and which are not?
Anyway, You need to ask developers before removing these indexes.

Searching over Oracle partitions

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.

Best way to identify a handful of records expected to have a flag set to TRUE

I have a table that I expect to get 7 million records a month on a pretty wide table. A small portion of these records are expected to be flagged as "problem" records.
What is the best way to implement the table to locate these records in an efficient way?
I'm new to Oracle, but is a materialized view an valid option? Are there such things in Oracle such as indexed views or is this potentially really the same thing?
Most of the reporting is by month, so partitioning by month seems like an option, but a "problem" record may be lingering for several months theorectically. Otherwise, the reporting shuold be mostly for the current month. Would you expect that querying across all month partitions to locate any problem record would cause significant performance issues compared to usinga single table?
Your general thoughts of where to start would be appreciated. I realize I need to read up and I'll do that but I wanted to get the community thought first to make sure I read the right stuff.
One more thought: The primary key is a GUID varchar2(36). In order of magnitude, how much of a performance hit would you expect this to be relative to using a NUMBER data type PK? This worries me but it is out of my control.
It depends what you mean by "flagged", but it sounds to me like you would benefit from a simple index, function based index, or an indexed virtual column.
In all cases you should be careful to ensure that all the index columns are NULL for rows that do not need to be flagged. This way your index will contain only the rows that are flagged (Oracle does not - by default - index rows in B-Tree indexes where all index column values are NULL).
Your primary key being a VARCHAR2 GUID should make no difference, at least with regards to the specific flagging of rows in this question, indexes will point to rows via Oracle internal ROWIDs.
Indexes support partitioning, so if your data is already partitioned, your index could be set to match.
Simple column index method
If you can dictate how the flagging works, or the column already exists, then I would simply add an index to it like so:
CREATE INDEX my_table_problems_idx ON my_table (problem_flag)
/
Function-based index method
If the data model is fixed / there is no flag column, then you can create a function-based index assuming that you have all the information you need in the target table. For example:
CREATE INDEX my_table_problems_fnidx ON my_table (
CASE
WHEN amount > 100 THEN 'Y'
ELSE NULL
END
)
/
Now if you use the same logic in your SELECT statement, you should find that it uses the index to efficiently match rows.
SELECT *
FROM my_table
WHERE CASE
WHEN amount > 100 THEN 'Y'
ELSE NULL
END IS NOT NULL
/
This is a bit clunky though, and it requires you to use the same logic in queries as the index definition. Not great. You could use a view to mask this, but you're still duplicating logic in at least two places.
Indexed virtual column
In my opinion, this is the best way to do it if you are computing the value dynamically (available from 11g onwards):
ALTER TABLE my_table
ADD virtual_problem_flag VARCHAR2(1) AS (
CASE
WHEN amount > 100 THEN 'Y'
ELSE NULL
END
)
/
CREATE INDEX my_table_problems_idx ON my_table (virtual_problem_flag)
/
Now you can just query the virtual column as if it were a real column, i.e.
SELECT *
FROM my_table
WHERE virtual_problem_flag = 'Y'
/
This will use the index and puts the function-based logic into a single place.
Create a new table with just the pks of the problem rows.

How to generate PHYSICALLY sorted GUID as primary key in Oracle?

I'm trying to generate a sorted GUID as a primary key in Oracle. In SQL Server I could use one of the following to sort the rows physically
By clustered primary key as a unique identifier.
By NEWSEQUENTIALID.
I have searched for an Oracle equivalent but failed to find a solution. I know about Is there a way to create an auto-incrementing Guid Primary Key in an Oracle database?, but there's no indication whether SYS_GUID() is sorted.
How could I create a sequential primary key in Oracle?
If you want to create a GUID then SYS_GUID() is what you should be using, you can create this in a table as per the linked question. It's unclear from the documentation whether SYS_GUID() is incrementing. It might be but that's not really a statement that imparts trust.
The next part of your question (and some comments) keeps asking about clustered primary keys. This concept does not exist in the same way in Oracle as it does in SQL Server and Sybase. Oracle does have indexed organized tables (IOT) ...
... a table stored in a variation of a B-tree index structure... rows
are stored in an index defined on the primary key for the table. Each
index entry in the B-tree also stores the non-key column values. Thus,
the index is the data, and the data is the index.
There are plenty of uses for IOTs but it's worth bearing in mind you're altering the physical structure of the database on the disk for "performance reasons". You're doing the ultimate of all premature optimizations using something that has both negative and positive aspects. Read the documentation and be sure that this is what you want to do.
I would generally use an IOT only when you don't care about DML performance but when you do a lot of range scans, or you need to order by the primary key. You create an IOT in the same way as you would an ordinary table, but because everything you want is now part of the table everything goes in your table definition:
create table test_table (
id raw(32) default sys_guid()
, a_col varchar2(50)
, constraint pk_my_iot primary key (id)
) organization index;
It's worth noting that even with an IOT you must use an explicit ORDER BY in order to guarantee returned order. However, because of the way this is stored Oracle can table a few short cuts:
select *
from ( select *
from test_table
order by id )
where rownum < 2
SQL Fiddle.
As with everything, test, don't assume that this is the structure you want.
Oracle has as SYS_GUID() function, which generates a 16-byte RAW datatype. But, I'm not sure what you mean by "sorted GUID". Can you elaborate?
Do you mean you need each generated GUID to sort "after" the previously generated GUID? I looked at the SYS_GUID() function, and it seems to generate GUIDs in sorted order, but looking at the documentation, I don't see anything that says that is guaranteed.
If I understand your question correctly, I'm not sure it's possible.
You may be able to use SYS_GUID() and prepend a sequence, to get your desired sort order?
Can you explain more about your use case?
Adding the following in response to comment:
Ok, now I think I understand. What I think you want, is something called an IOT, or Index Organized Table, in Oracle. It's a table that has an index strucure, and all data is clustered, or grouped by the primary key. More information is available here:
http://docs.oracle.com/cd/E16655_01/server.121/e17633/indexiot.htm#CNCPT721
I think that should do what you want.

Oracle: Index organised table with null values

I have a table which is basically a tree structure with a column parent_id and id.
parent_id is null for root nodes.
There is also a self referential foreign key, so that every parent_id has a corresponding id.
This table is mainly read-only with mostly infrequent batch updates.
One of the most common queries from the application which accesses this table is select ... where parent_id = X. I thought this might be faster if this table was index organised on parent_id.
However, I'm not sure how to index organise this table if parent_id can be null. I'd rather not fudge things so that parent_id=0 is some special id, as I'd have to add dummy values to the table to ensure the foreign key constraints are satisfied, and it also changes the application logic.
Is there any way to index organise a table by possible null value columns?
Solution from question asker:
I found I could get the same benefits from index organisation just by adding the queried columns to the end of the parent_id index, i.e. instead of:
create index foo_idx on foo_tab(parent_id);
I do:
create index foo_idx on foo_tab(parent_id, col1, col2, col3);
Where col1, col2, col3 etc are frequently accessed columns.
I've only done this with indexes which are used to return multiple rows which benefit from the ordering and hence disk locality provided by the index, instead of having to jump around the table. Indexes which are generally used to return single rows I've left to reference the table, as there is only one row to read anyway so locality matters much less.
Like I mentioned, this is a mainly read table, and also space is not a huge concern, so I don't think the overhead to writes caused by these indexes is a big concern.
(I realise this won't index null parent_ids, but instead I've made another index on decode(parent_id, null, 1, null) which indexes nulls and only nulls).
I would try adding the index on the single column parent_id.
If all of the columns in your index are non-null, then this row does not appear in your index.
So for the parent_id = X you cite above, this should use the index. However, if you're doing parent_id is null, then it won't use the index, and you'll be getting the same performance as you have now. This sounds like behaviour that would suit you.
I have used this in the past to improve the performance of queries. It works particulalry well if the number of items in the index is small compared to the number of rows in the database. We had about 3% of our rows in this particular index, and it flew :-)
But, as always, you need to try it and measure the difference in performance. Your mileage may vary.

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