Populate a Drop down box from a mySQL table in PHP

Since mysql_connect has been deprecated, connect and query instead with mysqli:

$mysqli = new mysqli("hostname","username","password","database_name");
$sqlSelect="SELECT your_fieldname FROM your_table";
$result = $mysqli -> query ($sqlSelect);

And then, if you have more than one option list with the same values on the same page, put the values in an array:

while ($row = mysqli_fetch_array($result)) {
    $rows[] = $row;
}

And then you can loop the array multiple times on the same page:

foreach ($rows as $row) {
    print "<option value='" . $row['your_fieldname'] . "'>" . $row['your_fieldname'] . "</option>";
}

Below is the code for drop down using MySql and PHP:

<?
$sql="Select PcID from PC"
$q=mysql_query($sql)
echo "<select name=\"pcid\">"; 
echo "<option size =30 ></option>";
while($row = mysql_fetch_array($q)) 
{        
echo "<option value='".$row['PcID']."'>".$row['PcID']."</option>"; 
}
echo "</select>";
?>

You will need to make sure that if you're using a test environment like WAMP set your username as root. Here is an example which connects to a MySQL database, issues your query, and outputs <option> tags for a <select> box from each row in the table.

<?php

mysql_connect('hostname', 'username', 'password');
mysql_select_db('database-name');

$sql = "SELECT PcID FROM PC";
$result = mysql_query($sql);

echo "<select name='PcID'>";
while ($row = mysql_fetch_array($result)) {
    echo "<option value='" . $row['PcID'] . "'>" . $row['PcID'] . "</option>";
}
echo "</select>";

?>