SQL query into option form - php

Question

I have a PHP code to get info from my Microsoft SQL server 2014, but it isnt working, the page it self works fine since it pops up as it should when i comment out the PHP code, but as soon as the PHP code isnt commented out, its just all white, so im assuming problem with the PHP code. I have to get the results from the query out into a drop down menu.
i use this code:
$servername = "VCCSQL03";
$username = "forecast";
$password = "Telefon2";
$dbname = "Forecast";
$connectionInfo = array("Database"=>$dbname, "UID"=>$username, "PWD"=>$password);
$conn = sqlsrv_connect($serverName, $connectionInfo);
if(!$conn) {
echo "Connection could not be established.<br />";
die( print_r( sqlsrv_errors(), true));
}
// Check connection
$result = sqlsrv_query($conn,"SELECT * FROM dbo.vw_BrandProduct");
if ($result->num_rows > 0) {
// output data of each row
while($row = sqlsrv_fetch_array($result)) {
echo "<option value='".$row['Brand_ProductID']."' name='".$row['Brand_ProductName']."'</option>";
}
} else {
echo "";
}
sqlsrv_close();

Answer 1

First and foremost, you do not have an open and closed select tag, and your option tags was missing a > to close it properly. Try the below revision, assuming connection is established on the page properly then this should work.
$connectionInfo = array( "Database"=>$dbname, "UID"=>$username, "PWD"=>$password);
$conn = sqlsrv_connect( $serverName, $connectionInfo);
if(!$conn) {
//// Check connection
echo "Connection could not be established.<br />";
die( print_r( sqlsrv_errors(), true));
}
$result = sqlsrv_query($conn,"SELECT * FROM dbo.vw_BrandProduct");
if ($result->num_rows > 0) {
// output data of each row
echo "<select name='products'>";
while($row = sqlsrv_fetch_array($result)) {
echo "<option value='".$row['Brand_ProductID']."'>$row['Brand_ProductName']</option>";
}
echo "</select>";
} else {
echo ""; } sqlsrv_close(); ?>