I am using the below query to get Ids like 12766% and 39998%. How I can use regexp_like to check 12%66 and 39%98?
select * from table t where regexp_like( UPPER(ID),'12766|39998')
You may use the following regex pattern:
^(12[0-9]*66|39[0-9]*98)$
Oracle query:
SELECT *
FROM yourTable
WHERE REGEXP_LIKE(ID, '^(12[0-9]*66|39[0-9]*98)$');
Demo
But actually, you might just want to use LIKE here:
SELECT *
FROM yourTable
WHERE ID LIKE '12%66' OR ID LIKE '39%98';
This would work find, so long as you don't mind exactly what comes in between the digits.
I found solution for this. We can use the below query for a%d to match strings like abcd,acd,aabcd etc. A period character (.) is a perfect replacement for % in regexp which can support one or more occurrence of any characters supported in database.
select * from table where REGEXP_LIKE (UPPER (ID), '^12.66.|EY39.98.')
Related
select * from user_table where name in ('123%','test%','dummy%')
How to ensure that this where clause is not an exact match, but a like condition?
In Oracle you can use regexp_like as follows:
select *
from user_table
where regexp_like (name, '^(123|test|dummy)')
The caret (^) requires that the match is at the start of name, and the pipe | acts as an OR.
Be careful though, because with regular expressions you almost certainly lose the benefit of an index that might exist on name.
Use like this,
select *
from user_table
where name LIKE '123%'
OR name LIKE 'test%'
OR name Like 'dummy%';
another option in MySQL
select * from user_table where name REGEXP '^123|^test|^dummy';
To not lose indexed access to rows in Oracle a table collection expression can be used:
SELECT
*
FROM
user_table
JOIN (SELECT column_value filter
FROM table(sys.odcivarchar2list('dummy%', '123%', 'test%'))
) ON user_table.name LIKE filter
The filter expressions must be distinct otherwise you get the same rows from user_table multiple times.
SQL newbie here, but I can't find the solution to something that looks easy:
The following query does not seem to have a valid syntax (ORA-00904: invalid identifier), but its logic should be clear. How can I achieve this in a query that needs to be speedy?
SELECT * FROM table WHERE LEFT(column,4)="abcd"
For this purpose, you should use like rather than left(). First, Oracle doesn't support left() (you need substr() instead). Second, like can make use of indexes because the wildcard is not at the beginning of the string:
SELECT *
FROM table
WHERE column like 'abcd%';
Oracle and some other products have substr.
SELECT * FROM tablename WHERE substr(columnname, 1, 4) = 'abcd'
I.e. single quotes for string literals!
ANSI SQL has substring:
SELECT * FROM tablename WHERE substring(columnname from 1 for 4) = 'abcd'
And others have left:
SELECT * FROM tablename WHERE LEFT(columnname,4) = 'abcd'
I am in a situation where I need to return results if some conditions on the string/character are met.
For example: to return only the names that contain 'F' character from the Person table.
How to create an SQL query based on such conditions? Is there any link to a documentation that explains how can SQL perform such queries?
Thanks in advance
The most basic approach is to use LIKE operator:
-- name starts with 'F'
SELECT * FROM person WHERE name LIKE 'F%'
-- name contains 'F'
SELECT * FROM person WHERE name LIKE '%F%'
(% is a wildcard)
Most RDBMS offer string operations which are able to perform that required task in one way or the other.
In MySQL you might use INSTR:
SELECT *
FROM yourtable
WHERE INSTR(Person, 'F') > 0;
In Oracle, this can be done, too.
In PostgreSQL, you can use STRPOS:
SELECT *
FROM yourtable
WHERE strpos(Person, 'F') > 0;
Usually there are several approaches to solve this, many would choose the LIKE operator. For more details, please refer to the documentation of the RDBMS of your choice.
Update
As requested by the questioner a few words about the LIKE operator, which are used not only in MySQL or Oracle, but in other RDBMS, too.
The use of LIKE will in some cases make your RDBMS try to use an index, it usually does not not try to do so if you use a string functions.
Example:
SELECT *
FROM yourtable
WHERE Person LIKE 'F%';
The query may look like this:
SELECT * FROM Person WHERE FirstName LIKE '%F%' OR LastName LIKE '%F%'
Is there any shorter way to look for multiple matches:
SELECT *
from table
WHERE column LIKE "AAA%"
OR column LIKE "BBB%"
OR column LIKE "CCC%"
This questions applies to PostgreSQL 9.1, but if there is a generic solution it would be even better.
Perhaps using SIMILAR TO would work ?
SELECT * from table WHERE column SIMILAR TO '(AAA|BBB|CCC)%';
Use LIKE ANY(ARRAY['AAA%', 'BBB%', 'CCC%']) as per this cool trick #maniek showed earlier today.
Using array or set comparisons:
create table t (str text);
insert into t values ('AAA'), ('BBB'), ('DDD999YYY'), ('DDD099YYY');
select str from t
where str like any ('{"AAA%", "BBB%", "CCC%"}');
select str from t
where str like any (values('AAA%'), ('BBB%'), ('CCC%'));
It is also possible to do an AND which would not be easy with a regex if it were to match any order:
select str from t
where str like all ('{"%999%", "DDD%"}');
select str from t
where str like all (values('%999%'), ('DDD%'));
You can use regular expression operator (~), separated by (|) as described in Pattern Matching
select column_a from table where column_a ~* 'aaa|bbb|ccc'
Following query helped me. Instead of using LIKE, you can use ~*.
select id, name from hosts where name ~* 'julia|lena|jack';
You might be able to use IN, if you don't actually need wildcards.
SELECT *
from table
WHERE column IN ('AAA', 'BBB', 'CCC')
I'm looking for a character that I can use in an Oracle contains to get ALL results.
If I search for the string "test" in a title column I use this statement:
select *
from my_table
where contains (title,
'<query><textquery grammar="CTXCAT">test</textquery></query>') > 0
With this statement I get the rows which have the "test"-string included in title-column.
BUT: Is there any way to use contains and select ALL rows?
Would this work for you?
select * from my_table
where title like
'<query><textquery grammar="CTXCAT">%</textquery></query>'
This uses the LIKE syntax and the % wildcard to achieve what I think you want.
The documentation says wildcards with the Oracle Text CONTAINS() predicate are % and _ just like the LIKE predicate.
Does the following work? I don't have an instance of Oracle handy to test.
select *
from my_table
where contains (title,
'<query><textquery grammar="CTXCAT">%</textquery></query>') > 0