What will be the output of the following MySQL statement?

SELECT *
FROM employee
WHERE lname LIKE ‘_a%e%’;

Solution (By Examveda Team)

This question is about using the LIKE operator in MySQL to filter data based on patterns in a column. Here's a breakdown of the code and the options:
The Code
```sql SELECT * FROM employee WHERE lname LIKE ‘_a%e%’; ```
* SELECT *: This tells MySQL to retrieve all columns (fields) from the table. * FROM employee: This specifies the table named "employee" to pull data from. * WHERE lname LIKE ‘_a%e%’: This is the filtering condition. Let's understand the pattern: * _a: This means the second letter in the last name must be "a". * %: This is a wildcard character representing any number of characters (zero or more). * e%: This means there must be at least one "e" somewhere after the second letter.
The Options
* Option A: This is partially correct. It correctly identifies that the second letter must be "a". However, it doesn't include the requirement for at least one "e" in the name. * Option B: This is incorrect. While the code does include an "e", it doesn't specify that the "e" must be anywhere in the name. The code is more specific. * Option C: This is correct. This option accurately reflects the pattern used in the LIKE clause: the second letter must be "a", and there has to be at least one "e" in the name. * Option D: This is incorrect. Only Option C is entirely accurate, and the other options are partially or fully wrong.
Therefore, the correct answer is Option C.