pdo 包装器最后一个 ID 返回 0

Question

hello Iam trying to make a project using php mvc and I used on it pdo wrapper package the main problem that when I use lastinsertid it always return 0 I know that I should use the same connection but in my case with the package how can I do it??

this is the model code

<?php
   namespace MVC\core;
    use Dcblogdev\PdoWrapper\Database as Database;

    class model{
   static function db(){
    $options = [
    //required
    'username' => 'root',
    'database' => 'gallery',
    //optional
    'password' => '',
    'type' => 'mysql',
    'charset' => 'utf8',
    'host' => 'localhost',
    'port' => '3306'
     ];
      return  $db = new Database($options);
    }
    }
    ?>

and here where I use the pdo with my function

    <?php
namespace MVC\model;
use MVC\core\model;
use MVC\core\session;
class adminpost extends model {
function getCategory(){
    return  model::db()->rows("SELECT * FROM `category` WHERE id NOT in (SELECT parent_id from category)");
}
function getposts(){
    return  model::db()->rows("select * from posts");
}
function getpostsByCategory($id){
    return  model::db()->rows("select * from posts where category_id = ?",[$id]);
}
function insertimage($data){
    return model::db()->insert('images', $data);

}
function insertpost($data){
     return model::db()->insert('posts', $data);

}
function lastid(){
    return model::db()->lastInsertId();

}
}


?>

and here is the controller where I use the model function

class adminpostcontroller extends controller{
function postinsert(){
$post = new adminpost;
$data = [
'title'=>$_POST['title'],
'text'=>$_POST['text']
];
$insert = $post->insertpost($data);
}
}

Answer 1

I'm assuming its that every time you call model::db() you're establishing a new connection due to new Database($options) , you need to create and retain one connection for the duration of your actions via a static reference or another most appropriate means.

A "cheap" way to do this is

static $db;
return isset( $db ) ? $db : $db = new Database($options);

Other ways could be to create a reference in your class like static $db; and use static::$db

class model{
  static $db;

followed by

return isset( static::$db ) ? static::$db : static::$db = new Database($options);