Question

的config.php

define('DB_TYPE', 'MYSQL');

dbManager.php

incude_once('config.php');
switch ( DB_TYPE ) {
   case 'MYSQL':    $this->_dataObject = MySqlDB::_getInstance();   break;
   case 'PGSQL':    $this->_dataObject = PostgreDB::_getInstance(); break;
   case 'SQLITE':   $this->_dataObject = SqliteDB::_getInstance();  break;
}

这段代码可以在没有案例的情况下重写吗？类似的东西：

$this->_dataObject = DB_TYPE::_getInstance();

这给了我一个错误Fatal error: Class 'DB_TYPE' not found .... in line ...

Answer 1

也许

$DBClassName = 'MySqlDB';
$this->_dataObject = $DBClassName::_getInstance();

Answer 2

尝试：

$c = TYPE;
$this->_dataObject = $c::_getInstance();

然而，这段代码看起来可能会受益于依赖注入

class YourClass {

  public function __construct(DataObject $dataObject) {
    $this->_dataObject = $dataObject;
  }

}

然后将数据对象传递给您的类，同时实例化它

Answer 3

$dbClasses = array(
    'MYSQL' => 'MySqlDB',
    'PGSQL' => 'PostgreDB',
    'SQLITE'=> 'SqliteDB',
};

$dbClass = $dbClasses[DB_TYPE];

$this->_dataObject = $dbClass::_getInstance();

从代码中删除案例

3 个答案: