I have one table in Oracle 11g database
I am storing some value like '|' separated mode as below
Table: ABC
FIELD: XYZ
Values :
9
1|12
52
5|112
Now I want to find ID from those values
I used REGEXP_LIKE but it will get all the rows containing value
Suppose I will search for 5 then it will give result for '5' and also for '52'
please help me to search particulate id from this field
Thanks
Try this.
select * from table_name
where '|' || column_name || '|' like '%|' || 'search_string' || '|%';
Sample here.
Related
i have a table that store mutiple values in column i need to find if the value exists then return entire row
Example : 963963963;35793579;741741
Query am using
select * from table where column like (%963%);
which return all the rows that have the value starts or ends with 963
i need only one row that match with 963
It is generally bad table design to store semicolon separated data this way. If you must continue with your current design, you may use the following query to find an exact match for 963:
SELECT *
FROM yourTable
WHERE ';' || col || ';' LIKE '%;963;%';
Am working with an Oracle 11g db in which I need to run a query that returns all rows with values in a specific format.
I.e., I need to find all rows with a value like 'abc1234'. The actual values aren't important....what I need is to find all rows with the first 3 characters being alpha and the subsequent 4 characters all being numeric.
Any input would be much appreciated.
Might not be exact since I don't have an Oracle server handy to test but something like this should get you started in the right direction:
SELECT * FROM your_table WHERE REGEXP_LIKE(your_column, '([a-z]\3[0123456789]\4', 'i')
This will check all rows in your table where the specified column has any alphabet character A to Z three times followed by 4 numbers and return that list. The 'i' at the end just says ignore case on the alphabet part. You can choose to not have that option if you so wish.
Here are a couple links as well with more information on the regexp_like syntax:
Oracle SQL - REGEXP_LIKE contains characters other than a-z or A-Z
https://docs.oracle.com/cd/B28359_01/server.111/b28286/conditions007.htm#SQLRF00501
Try it this will work definitely:
Let's take a sample example:
For example, create a sample table
CREATE TABLE bob (
empno VARCHAR2(10) NOT NULL,
ename VARCHAR2(15) NOT NULL,
ssn VARCHAR2(10) NOT NULL);
Insert data into that table
Insert into bob ( empno,ename,ssn) values ('abC0123458','zXe5378023','0pl4783202');
Insert into bob ( empno,ename,ssn) values ('0123abcdef','a12dldfddd','Rns0101010');
Query to select the all the columns
select * from bob
where length(trim(translate(substr(empno,1,3),'abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ',' '))) IS NULL
AND
length(trim(translate(substr(empno,4,4),'0123456789',' '))) IS NULL;
To do the same thing on multiple columns use union operator
select * from bob
where length(trim(translate(substr(empno,1,3),'abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ',' '))) IS NULL
AND
length(trim(translate(substr(empno,4,4),'0123456789',' '))) IS NULL;
union
select * from bob
where length(trim(translate(substr(ename,1,3),'abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ',' '))) IS NULL
AND
length(trim(translate(substr(ename,4,4),'0123456789',' '))) IS NULL;
I have a table structure like below:
TestTable:
ID(number)|Names(clob)
1 'a','b','c','d'
2 'b','c','d'
3 'g','h','e'
Now I want to select rows in which Names column contains 'b',so the expected output will be the first 2 columns.
How can do it in Oracle.
Please remember, the solution has to work in Oracle 10g as well as 11g.
Thanks in advance.
Try this using LIKE operator:
select
*
from your_table
where ',' || Names || ',' like '%,b,%';
Concatenated commas are to cover for the cases where b is present at the start or end of the string.
How to generate hash key using multiple columns in oracle? Could you please give me an example?
Depending on what you want a hash for, you might use ORA_HASH:
SELECT ORA_HASH(name || '~' || surname || '~' || position) FROM emp;
http://docs.oracle.com/cd/B12037_01/server.101/b10759/functions097.htm
Or you might use DBMS_CRYPTO.hash:
http://docs.oracle.com/cd/E11882_01/appdev.112/e40758/d_crypto.htm#i1002022
Example if an ID is 1213 i want show **13.
If it's a number
select '**' || substr(to_char(id),3)
from my_table
Or, if it's already a character
select '**' || substr(id,3)
from my_table
This concatenates ** onto the beginning of the string, using the Oracle concatenation operator || and removes the first two characters of the id using substr.
Here's a SQL Fiddle to demonstrate.
If you don't want to sacrifice performance too much, to mask first two characters you can use-
SQL> select regexp_replace('1213','(.)2','**') from dual; --if VARCHAR
MASKED
------------
**13
SQL> select regexp_replace(1213,'(.)2','**') from dual; --if NUMBER
MASKED
------------
**13
REGEXP_REPLACE will work alike on NUMBER and VARCHAR so you save some conversion time there.
Consecutively, you can create a Function Based Index on the regexp function operation to optimize the query like (considering you would always want to mask only first two characters of ID) -
CREATE INDEX
mask_id
ON
table_name
(regexp_replace(id,'(.)2','**'));