<?PHP
$dbhost = 'localhost';
$dbuser = 'user';
$dbpass = 'password';
$db = new MysqLi($dbhost,$dbuser,$dbpass,'images_db');
if($db->connect_errno > 0){
die('Unable to connect to database [' . $db->connect_error . ']');
}
else{
echo "Connected to database";
}
//filename,mime_tyPE and file_size are columns in the table images
$stmt = $db->PRepare("INSERT INTO images (filename,mime_type,file_size) VALUES (?,?,?)");
$string1 = 'string 1';
$string2 = 'string 2';
$stmt->bind_param('ssi',$string1,$string2,123);
$stmt->execute();
$stmt->close();
$MysqLi->close();
?>

当我执行代码时,没有任何东西被添加到MysqL数据库中.但是当我评论出这条线时

$stmt->bind_param('ssi',123);

并将字符串和整数值直接插入$db-> prepare语句(替换问号),这一切都很好用,并将行添加到数据库表中.

我在bind_param行中做错了什么,阻止将新行添加到数据库中？