The following query selects details of all those employees whose name starts with 'K'.

Worked Examples · Example 9.13

Q. The following query selects details of all those employees whose name starts with 'K'.

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Example 9.13 asks for employees whose name starts with 'K'. This is substring pattern matching, done with the LIKE operator and the % wildcard — not a set-difference operation.

On the real EMPLOYEE table (Table 9.8) of the OFFICE database, the query is:

SELECT * FROM EMPLOYEE WHERE Ename LIKE 'K%';

'K%' matches any name that begins with the literal character 'K', followed by zero or more characters. Checking every real name in the table — Aaliya, Kritika, Shabbir, Gurpreet, Joseph, Sanya, Vergese, Nachaobi, Daribha, Tanya — only Kritika begins with 'K'. …

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This is Example 9.13 from the OFFICE database's EMPLOYEE table (Table 9.8). The query uses LIKE 'K%' to find names starting with 'K'; only Kritika matches.

The real EMPLOYEE table (Table 9.8)

EmpNo	Ename	Salary	Bonus	DeptId
101	Aaliya	10000	234	D02
102	Kritika	60000	123	D01
103	Shabbir	45000	566	D01
104	Gurpreet	19000	565	D04
105	Joseph	34000	875	D03
106	Sanya	48000	695	D02
107	Vergese	15000	NULL	D01
108	Nachaobi	29000	NULL	D05
109	Daribha	42000	NULL	D04
110	Tanya	50000	467	D05

When you need to find data that matches a pattern rather than an exact value, SQL provides the LIKE operator together with wildcard characters:

% matches zero, one, or multiple characters.
_ matches exactly one character.

Here we want employees whose name starts with 'K', so the pattern is 'K%':

mysql> SELECT * FROM EMPLOYEE
    -> WHERE Ename like 'K%';
``` …

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Q.The following query selects details of all those employees whose name starts with 'K'.

Concept understanding — Set Difference

The real EMPLOYEE table (Table 9.8)

Q.The following query selects details of all those employees whose name starts with 'K'.

The real EMPLOYEE table (Table 9.8)

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