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SOUNDS LIKE Function in MySQL

Last Updated : 14 Dec, 2020
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SOUNDS LIKE :
This function in MySQL is used to compare the Soundex codes of a given two string expressions. It is used as SOUNDEX(expr1) = SOUNDEX(expr2) to retrieve strings that sound similar.

Syntax :

expr1 SOUNDS LIKE expr2

Parameter :
It accepts two parameter as mentioned above and described below.

  • expr1 – The first string which we want to compare.
  • expr2 –The second string which we want to compare.

Returns :
It compares the Soundex code of two string values and returns the output.

Example-1 :
Comparing similar two given string using SOUNDS LIKE Function.

SELECT 'geeks' SOUNDS LIKE 'geeks' 
as Result;

Output :

Result

    1

Example-2 :
Comparing similar two given strings using SOUNDS LIKE Function.

SELECT 'geeks' SOUNDS LIKE 'for' 
as Result;

Output :

Result

    0

Example-3 :
The following example shows returns all the rows which contain an Employee name whose first name sounds like ‘Sayan’.

CREATE TABLE Employee
(
Employee_id INT AUTO_INCREMENT,  
First_name VARCHAR(100) NOT NULL,
Last_name VARCHAR(100) NOT NULL,
Joining_Date DATE NOT NULL,
PRIMARY KEY(Employee_id )
);

Inserting some data to the Employee table :

INSERT INTO Employee
(First_name ,Last_name , Joining_Date )
VALUES
('Sayantan', 'Majumdar', '2000-01-11'),
('Anushka', 'Samanta', '2002-11-10' ),
('Sayan', 'Sharma', '2005-06-11' ),
('Shayari', 'Das', '2008-01-21' ),
('Sayani', 'Jain', '2008-02-01' ),
('Tapan', 'Samanta', '2010-01-11' ),
('Deepak', 'Sharma', '2014-12-01'  ),
('Ankana', 'Jana', '2018-08-17'),
('Shreya', 'Ghosh', '2020-09-10') ;

So, the Employee Table is as follows.

select * from Employee ;

Output :

Employee_id First_name Last_name Joining_Date
1 Sayantan Majumdar 2000-01-11
2 Anushka Samanta 2002-11-10
3 Sayan Sharma 2005-06-11
4 Shayari Das 2008-01-21
5 Sayani Jain 2008-02-01
6 Tapan Samanta 2010-01-11
7 Deepak Sharma 2014-12-01
8 Ankana Jana 2018-08-17
9 Shreya Ghosh 2020-09-10

Now, we are going to check those employee whose first name sounds like ‘sayan’.

SELECT * FROM Employee 
WHERE First_name SOUNDS LIKE 'Sayan' ;

Output :

Employee_id First_name Last_name Joining_Date
3 Sayan Sharma 2005-06-11
5 Sayani Jain 2008-02-01

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