在php函数中使用变量返回值

Question

I have a function that is supposed to give me a staff members name from his ID number so my php is like this

$staffId = $_GET['staff_id'];
$staff = staff_load($staffId);

and my function is like this

function staff_load()
{
$dbh = dbh_get(); //connects to database

$sql = 'select user_name from user_staff where user_id = ?';
$stmt = $dbh->prepare($sql);
$stmt->execute();
$staff = $stmt->fetch();

dbh_free($dbh); //disconnects from database
return $staff;
}

But when I try and use the $staff variable, it shows nothing. I can't work out what I'm doing wrong. I've tried a bunch of variants and gotten nowhere except frustrated.

<td>Book for ' . $staff . '</td>

Answer 1

First, you forgot to include the parameter in your function definition:

function staff_load($id) {

Then you need to bind the parameter of the prepared statement.

$sql = 'select user_name from user_staff where user_id = $1';
$stmt = $dbh->prepare($sql);
$stmt->execute(array($id));

Next, fetch() returns an array, you need to extract the user_name element from the array:

$row = $stmt->fetch();
if ($row) {
    $staff = $row['user_name'];
} else {
    $staff = false;
}