Populate drop down form SQL server using PHP and HTML - php

Question

Whenever I try to run the below code, I get the passed variable ". $record . " as it is.
I am trying to populate a drop-down from a database, but I keep getting a value which I am passing in the while loop.. I know I am making a simple mistake.Can someone please help to solve this issue
<select name="c" multiple="multiple">
<option>--select--</option>
<?php
$dbname = "abc";
$serverName = "xyz"; //serverName\instanceName
try
{
$connectionInfo = array( "Database"=>$dbname);
$conn = sqlsrv_connect( $serverName, $connectionInfo);
$SQL_string= "Select distinct [CC] from [abc].[dbo].
[abcdefg] ";
$stmt = sqlsrv_query($conn, $SQL_string);
$records[] = array();
while ($row = sqlsrv_fetch_array($stmt,
SQLSRV_FETCH_ASSOC))
{
$records[] = $row['CC'];
}
foreach ($records as $record)
{
echo' <option value="'. $record . '">' .$record.
'</option>';
}
if( $stmt === false ) {
die( print_r( sqlsrv_errors(), true));
}
}
catch(Exception $ex)
{
echo "Connection failed: " . $ex->getMessage();
}
?>

Answer 1

Here's an attempt at simplifying this a bit. I can't guarantee it will work, but it may bring to light the error you are facing:
<select name="c" multiple="multiple">
<option>--select--</option>
<?php
$dbname = "abc";
$serverName = "xyz"; //serverName\instanceName
try
{
$connectionInfo = array( "Database"=>$dbname);
$conn = sqlsrv_connect( $serverName, $connectionInfo);
if( $conn === false ) {
die( print_r( sqlsrv_errors(), true));
}
$SQL_string= "Select distinct [CC] from [abc].[dbo].[abcdefg];";
$stmt = sqlsrv_query($conn, $SQL_string);
if( $stmt === false ) {
die( print_r( sqlsrv_errors(), true));
}
while ($row = sqlsrv_fetch_array($stmt,SQLSRV_FETCH_ASSOC))
{
echo '<option value="'.$row['CC'].'">'.$row['CC'].'</option>';
}
sqlsrv_free_stmt( $stmt);
}
catch(Exception $ex)
{
echo "Caught exception error: " . $ex->getMessage();
}
?>
</select>