The following query selects names of all employees containing 'se' as a substring in name.

Worked Examples · Example 9.16

Q. The following query selects names of all employees containing 'se' as a substring in name.

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Example 9.16 needs the LIKE operator with % on both sides of 'se' to find it anywhere in the name — not the set-difference EXCEPT operator.

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

SELECT Ename FROM EMPLOYEE WHERE Ename LIKE '%se%';

'%se%' means: any characters (or none), then the literal 'se', then any characters (or none) — i.e. 'se' can appear anywhere in the name. Checking the ten real names — Aaliya, Kritika, Shabbir, Gurpreet, Joseph, Sanya, Vergese, Nachaobi, Daribha, Tanya — 'se' occurs inside Joseph and Vergese only. …

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Example 9.16, on the real EMPLOYEE table (Table 9.8): Ename LIKE '%se%' finds names containing 'se' anywhere — Joseph and Vergese.

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

To find a substring anywhere within a name, % is placed on both sides of the target text:

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

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Q.The following query selects names of all employees containing 'se' as a substring in name.

Concept understanding — Set Difference

The real EMPLOYEE table (Table 9.8)

Q.The following query selects names of all employees containing 'se' as a substring in name.

The real EMPLOYEE table (Table 9.8)

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