[mysql] Select all where [first letter starts with B]

This is a follow up question to my previous one. I want to write a MYSQL statement that echoes every entry that starts with letter B.

Function.php

function getCategory() {
$query = mysql_query("SELECT author FROM lyrics WHERE author [starts with letter B]") or die(mysql_error());
while ($row = mysql_fetch_assoc($query)) { ?>
    <p><a href="##"><?= $row['author']; ?></a></p>
    <?php }

Category.php?category=b

<?php include 'includes/header.php' ?>
<?php getCategory(); ?>
<?php include 'includes/footer.php' ?>

Like that I guess. And then one for every letter of the alphabet, and one with misc (numbers etc)

Following your comment posted to ceejayoz's answer, two things are messed up a litte:

1. $first is not an array, it's a string. Replace $first = $first[0] . "%" by $first .= "%". Just for simplicity. (PHP string operators)

2. The string being compared with LIKE operator should be quoted. Replace LIKE ".$first."") by LIKE '".$first."'"). (MySQL String Comparison Functions)

This will work for MYSQL

SELECT Name FROM Employees WHERE Name REGEXP '^[B].*$'

In this REGEXP stands for regular expression

and

this is for T-SQL

SELECT Name FROM Employees WHERE Name LIKE '[B]%'

SQL Statement:

 SELECT * FROM employee WHERE employeeName LIKE 'A%';

Result:

Number of Records: 4

employeeID  employeeName    employeeName    Address City    PostalCode  Country

1           Alam             Wipro          Delhi   Delhi   11005      India

2           Aditya           Wipro          Delhi   Delhi   11005      India

3           Alok             HCL            Delhi   Delhi   11005      India

4           Ashok            IBM            Delhi   Delhi   11005      India

You can use:

WHERE LEFT (name_field, 1) = 'B';