从表中选择id = name？ ...查询失败

Question

I have a MySQL table called Users. The rows have 4 fields and look like:

  INT(10) |  serialized php array    |    CHAR(24) |  CHAR(254)
   id     |   profile                | name        |  email
   1      | ..ybl4JFtmhQEQVwQOXjo... | TestName    |  [e-mail]
   2      | ..hd+F2yZVYPlxmyalrQo... | NameTest    |  [e-mail]

I have a name of a user (TestName) and want to get the unique user id (1). Why does following code not work?

function dataBase() {
    $dataBase = @new MySQLi("localhost", "name", "password", "databasename");
    if (mysqli_connect_errno()) {
        safeExit("Failed to establish connection to MySQL database, error: ".mysqli_connect_error(), 'msgError');
    } else {
        return $dataBase;
    }
}

// connect to database
$db = dataBase();

$sql = 'SELECT
            id
        FROM
            Users
        WHERE
            name = ?
        LIMIT
            1';
$stmt = $db->prepare($sql);
if (!$stmt) {
    safeExit($db->error, 'msgError');
}

$name = 'TestName';

$stmt->bind_param('s', $name);
if (!$stmt->execute()) {
    safeExit($stmt->error, 'msgError');
}
    $stmt->bind_result($uid);
    $stmt->close();
echo $uid;

This code outputs a zer0. I can exclude syntax errors. Something must be wrong with the query.

Answer 1

You have not fetched the row from $stmt. Look over the examples on the mysqli::bind_result() documentation to see how this is usually done with a call to $stmt->fetch() either singly when expecting one row, or in a while loop.

$stmt->bind_param('s', $name);
if (!$stmt->execute()) {
    safeExit($stmt->error, 'msgError');
}

$stmt->bind_result($uid);

// Fetch the row so it populates the result variable
$stmt->fetch();
echo $uid;

$stmt->close();