SQL Server - LIKE Clause
The SQL Server (Transact-SQL) LIKE clause is used in a WHERE clause to search for a specified pattern in a specified column. The wildcards which are used in conjunction with the LIKE clause are given below:
Note: The NOT LIKE clause is the negation of LIKE clause.
Wildcard Characters in SQL Server
Symbol | Description | Example |
---|---|---|
% | Represents zero, one or multiple characters. | 'J%' represents a value that start with "J", for example - John, Jo and Jack etc. |
_ | Represents one character. | '_o%' represents a value that have "o" in the second position, for example - John, Jo and Journey etc. |
[] | Represents any single character specified within []. | '[ack]%' represents a value that starts with a, c or k, for example - ant, cat or kite etc. |
- | Represents a range of characters. | '[a-d]%' represents a value that starts with a, b, c or d, for example - ant, bat, cat or dog etc. |
[^] | Represents any character not specified within []. | '[^ack]%' represents a value that does not start with a, c or k, for example - bat, dog or rat etc. |
Syntax
The syntax for using LIKE Clause in SQL Server (Transact-SQL) is given below:
SELECT column1, column2, ... FROM table_name WHERE column LIKE pattern;
The table below describes patterns which is used with LIKE clause and uses (%) and (_).
Pattern | Description |
---|---|
'J%' | A value that start with "J". |
'%n' | A value that end with "n". |
'%oh%' | A value that have "oh" in any position. |
'_o%' | A value that have "o" in the second position. |
'J_%' | A value that start with "J" and have at least 2 characters. |
'J__%' | A value that start with "J" and have at least 3 characters. |
'J%n' | A value that start with "J" and ends with "n". |
Example:
Consider a database containing a table called Employee with the following records:
EmpID | Name | City | Age | Salary |
---|---|---|---|---|
1 | John | London | 25 | 3000 |
2 | Marry | New York | 24 | 2750 |
3 | Jo | Paris | 27 | 2800 |
4 | Kim | Amsterdam | 30 | 3100 |
5 | Ramesh | New Delhi | 28 | 3000 |
6 | Huang | Beijing | 28 | 2800 |
-
Using the % Wildcard : To select all records of the Employee table with Name starting with 'Jo', the query is given below.
SELECT * FROM Employee WHERE Name LIKE 'Jo%';
This will produce the result as shown below:
EmpID Name City Age Salary 1 John London 25 3000 3 Jo Paris 27 2800 -
Using the % Wildcard with NOT LIKE operator: NOT LIKE operator is used as the negation of LIKE operator. For example, to select all records of the Employee table with Name not starting with 'Jo', the following query can be used:
SELECT * FROM Employee WHERE Name NOT LIKE 'Jo%';
This will produce the result as shown below:
EmpID Name City Age Salary 2 Marry New York 24 2750 4 Kim Amsterdam 30 3100 5 Ramesh New Delhi 28 3000 6 Huang Beijing 28 2800 -
Using the _ Wildcard : To select all records of the Employee table with Name containing 'o' as second character, the query is mentioned below.
SELECT * FROM Employee WHERE Name LIKE '_o%';
The result of the above code will be:
EmpID Name City Age Salary 1 John London 25 3000 3 Jo Paris 27 2800 -
Using the [] Wildcard : To select all records of the Employee table with Name starting with 'K', 'R' or 'M', the query is:
SELECT * FROM Employee WHERE Name LIKE '[KRM]%';
The result of the code will be:
EmpID Name City Age Salary 2 Marry New York 24 2750 4 Kim Amsterdam 30 3100 5 Ramesh New Delhi 28 3000 -
Using the ^ Wildcard : To select all records of the Employee table with Name not starting with 'K', 'R' or 'M', the query is:
SELECT * FROM Employee WHERE Name LIKE '[^KRM]%';
The result of the code will be:
EmpID Name City Age Salary 1 John London 25 3000 3 Jo Paris 27 2800 6 Huang Beijing 28 2800 -
Using the - Wildcard : To select all records of the Employee table with Name starting with 'G' to 'J', the query is:
SELECT * FROM Employee WHERE Name LIKE '[G-J]%';
The result produced by the code will be following:
EmpID Name City Age Salary 1 John London 25 3000 3 Jo Paris 27 2800 6 Huang Beijing 28 2800