I have a large table of 7 column in Oracle 11G. Total size of the table is more than 3GB and total row in this table is 1876823. Query we are using
select doc_mstr_id from index_mstr where page_con1 like('%sachin%') it is taking almost a minute. please help me to optimize the query as well as proper indexing for this table. Please let me know if partitioned is required for this table.
Below are the column description
INDEX_MSTR_ID NUMBER
DOC_MSTR_ID NUMBER
PAGE_NO NUMBER
PAGE_PART NUMBER
PAGE_CON1 VARCHAR2(4000)
FILE_MODIFIED_DATE DATE
CREATED_DATE DATE
This query is always going to result in a full table scan. Your only filter cannot use a B-TREE index, due to the leading wildcard:
where page_con1 like('%sachin%')
If you want to do lots of queries of this nature you need to build a Text index on that column. From its datatype page_con1 appears to hold text fragments rather than full documents so you should use a CTXCAT index. This type of index has the advantage of being transactional, rather than requiring background maintenance. Find out more.
Your query would then look like this:
select doc_mstr_id from index_mstr
WHERE CATSEARCH(page_con1, 'sachin') > 0;
Related
We have a table that doesn't have much data. The table has 3 partitions and we are deleting data in one partition only.
delete from table AB partition(A) where id=value;
here id has an index also but still delete is slow.
The datatype of id is varchar2 and the value is number.
Please help me to understand why the delete statement is slow.
I don't think the index has much use in this case. It has to evaluate every single row in the partition to see if it matches id=value. Typically this will be a full table scan and no index will be used. It totally depends on the number of rows in the partition how long it will take. But maybe i did not understand the question properly. I presumed "value" is a column in the same table, like ID.
Which of the below would be the best way to create an Index, so that my query gives faster results?
My query is:
select emp_name,emp_last_name, salary
from employees_table
where salary <=2000;
1. create index emp_index on employees_table (salary);
2. create index emp_index on employees_table (emp_name,emp_last_name,salary);
In principle create index emp_index on employees_table (emp_name,emp_last_name,salary); would be the better one, because for your query Oracle has to read only the index but not any table data.
However, this means you create a separate index for each particular query which is certainly an overkill.
For 2 index Oracle may will be select table access full, if disctinct values in firsts columns of index to much, Oracle give you Table access full, if distinct values in first columns of index not much then you give index skip scan.
For first index you will see “index range scan” + “table access by rowid”....
but for both index’s - if sql returns >5-7% rows then Oracle maybe want to use “Table access full”...
All depends on your data in Table and count return rows
I have table A(id,name,code)
I have sql statement:
Select * from A where upper(code || name) like upper('%<search text>%');
How to create an index so that the following statement has an index?
Question for two option: table partitioned, and table not partitioned
Thanks & BR
Do you have a performance issue or is this just a hypothetical question?
An index is unlikely to help with this example: a full table scan will probably be the quickest solution. Why? Your table has 3 columns. The best index would be one that avoided looking in the table at all e.g.
create index ai on a (code, name, id);
But that needs to contain all the same data as the table plus a ROWID for each table row - so it is going to be bigger than the table and take longer to scan. You could try putting the columns in the index with the least selective first and using compression:
create index ai on a (code, name, id) compress;
Now the index may be smaller than the table - it depends on how selective the code and name columns are. If it is small enough, the optimizer might decide to use it instead of the table. It still contains all the IDs and ROWIDs so the reduction in size probably won't be dramatic. In the test case I set up the compressed index is about half the size of the table, yet Explain Plan shows the query has a higher cost if I use a hint to force it to use the index - maybe due to overheads of compression, I don't know.
You could look into Oracle Text and the CONTAINS expression - but then you would be writing a different query, not using LIKE.
I have a few tables with about 17M rows that all have a date column I would like to be able to utilize frequently for searches. I am considering either just throwing an index on the column and see how things go or sorting the items by date as a one time operation and then inserting everything into a new table so that the primary key ascends as the date ascends.
Since these are both pretty time consuming I thought it might be worth it to ask here first for input.
The end goal is for me to load sql queries into pandas for some analysis if that is relevant here.
The index on a date column makes sense when you are going to search the table for a given date(s), e.g.:
select * from test
where the_date = '2016-01-01';
-- or
select * from test
where the_date between '2016-01-01' and '2016-01-31';
-- etc
In these queries there is no matter whether the sort order of primary key and the date column are the same or not. Hence rewriting the data to the new table will be useless. Just create an index.
However, if you are going to use the index only in ORDER BY:
select * from test
order by the_date;
then a primary key integer index may be significantly (2-4 times) faster then an index on a date column.
Postgres supports to some extend clustered indexes, which is what you suggest by removing and reinserting the data.
In fact, removing and reinserting the data in the order you want will not change the time the query takes. Postgres does not know the order of the data.
If you know that the table's data does not change. Then cluster the data based on the index you create.
This operation reorders the table based on the order in the index. It is very effective until you update the table. The syntax is:
CLUSTER tableName USING IndexName;
See the manual for details.
I also recommend you use
explain <query>;
to compare two queries, before and after an index. Or before and after clustering.
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.