How to write a secure SELECT query which has a variable number of user-supplied values with mysqli?

Question

I would like some help to execute a foreach into my code.

I post multiple value into username : as username[0] = user1 , user2

but my foreach gi

Answer 1

Remove the '$value = $username_grab;' inside the foreach

i.e.

foreach ($username_grab as $value){
    // $value = $username_grab;   This does not belong

If that doesn't help, can you show the code this POSTs from and do a

print_r($username_grab)

just after you set it

Edit

<?php
$companyname = $_POST['companyname'];
$username_grab = $_POST['username'];                            // This (based on your example) is a weird single index array - Array ( [0] => user1,user2 )
// $username = implode(",", $username_grab);                    // implode turns an array into a string using a delimiter
$usernames = explode(",", $username_grab[0]);                   // explode turns a string into an array by delimiter. The [0] index is for the post coming through as such
// $usernames should now look something like this. This is the array you will be working with
// Array ( [0] => user1, [1] => user2 )

foreach($usernames AS $value)
{
    $sql = "select * from linked_user where username = '" . mysqli_escape_string($value) . "' and company_name = '" . mysqli_escape_string($companyname) . "'"; // SQL injection is a threat. Seperate issue to look into   
    $res = mysqli_query($conn,$value);
    $cnt = 0;                                                   // Looking for multiple results. Counter for array
    while($row = mysqli_fetch_array($res))
    {
        $returnValue[$cnt]['username'] = $row['username'];      // Create a two dimensional array (multiple users. Each has their own index then
        $returnValue[$cnt]['user_scid'] = $row['user_scid'];    // Can replace the $cnt field with $row['userid'] (or whatever) if you have a primary key on the table i.e. $returnValue[$row['userid']]['username'] = $row['username'];
    }
    $cnt++;                                                     // Increase the counter for the next user
}

// $returnValue should now look something like this
// Array ( 
// [0] => Array (
//          ['username'] => 'user1', 
//          ['user_scid'] => 'whateverthedbsaysforuser1'),
// [1] => Array (
//          ['username'] => 'user2', 
//          ['user_scid'] => 'whateverthedbsaysforuser2'),
// )
echo json_encode($returnValue);
?>

Answer 2

The task you are to perform is a prepared statement with a variable number of placeholders. This is simpler in PDO, but I'll show you the mysqli object-oriented style approach. No matter what, always print a json encoded array so that your receiving script knows what kind of data type to expect.

I had a snippet laying around that includes a full battery of diagnostics and error checking. I have not tested this script, but it has quite the resemblance to this post of mine.

if (empty($_POST['companyname']) || empty($_POST['username'])) {  // perform any validations here before doing any other processing
    exit(json_encode([]));
}

$config = ['localhost', 'root', '', 'dbname'];  // your connection credentials or use an include file
$values = array_merge([$_POST['companyname']], explode(',', $_POST['username']));  // create 1-dim array of dynamic length
$count = sizeof($values);
$placeholders = implode(',', array_fill(0, $count - 1, '?'));  // -1 because companyname placeholder is manually written into query
$param_types = str_repeat('s', $count);
if (!$conn = new mysqli(...$config)) {
    exit(json_encode("MySQL Connection Error: <b>Check config values</b>"));  // $conn->connect_error
}
if (!$stmt = $conn->prepare("SELECT user_scid, user_scid FROM linked_user WHERE company_name = ? AND username IN ({$placeholders})")) {
    exit(json_encode("MySQL Query Syntax Error: <b>Failed to prepare query</b>"));  // $conn->error
}
if (!$stmt->bind_param($param_types, ...$values)) {
    exit(json_encode("MySQL Query Syntax Error: <b>Failed to bind placeholders and data</b>"));  // $stmt->error;
}
if (!$stmt->execute()) {
    exit(json_encode("MySQL Query Syntax Error: <b>Execution of prepared statement failed.</b>"));  // $stmt->error;
}
if (!$result = $stmt->get_result()) {
    exit(json_encode("MySQL Query Syntax Error: <b>Get Result failed.</b>")); // $stmt->error;
}
exit(json_encode($result->fetch_all(MYSQLI_ASSOC)));

If you don't want the bloat of all of those diagnostic conditions and comments, here is the bare bones equivalent which should perform identically:

if (empty($_POST['companyname']) || empty($_POST['username'])) {
    exit(json_encode([]));
}

$values = explode(',', $_POST['username']);
$values[] = $_POST['companyname'];
$count = count($values);
$placeholders = implode(',', array_fill(0, $count - 1, '?'));
$param_types = str_repeat('s', $count);

$conn = new mysqli('localhost', 'root', '', 'dbname');
$stmt = $conn->prepare("SELECT user_scid, user_scid FROM linked_user WHERE username IN ({$placeholders}) AND company_name = ?");
$stmt->bind_param($param_types, ...$values);
$stmt->execute();
$result = $stmt->get_result();
exit(json_encode($result->fetch_all(MYSQLI_ASSOC)));