mysql_query() 实施结果一直为false

mysql_query() 执行结果一直为false
1）现有一个数据库名为test，里面只有一个表student。
属性名称：ID, Name, Email.
2）尝试着将数据库连接与操作封装成一个类DatabaseManager,并扩展了一个类StudentDetailsDataManager来获取学生信息。
3）问题：能够连接到test数据库，sql语句在数据库中测试过没有问题，但mysql_query()执行sql语句结果一直为false。不知什么问题？

代码如下：
数据库操作基类：DatabaseManager

PHP code

<!--Code highlighting produced by Actipro CodeHighlighter (freeware)http://www.CodeHighlighter.com/-->//DatabaseManager.php<?php class DatabaseManager{                protected $host;        protected $name;        protected $user;        protected $psw;                protected $connection;        protected $close_flag;                public function __construct($connection,$close_flag){            $this->connection = $connection;            $this->connection = $close_flag;        }                protected function db_open(){            if(empty($this->connection)){                $this->connection = mysql_connect($this->host,$this->user,$this->psw);                if (!$this->connection) {                    $this->db_handle_error_connetion();                    return false;                }                if (!mysql_select_db($this->name,$this->connection)) {                    $this->da_handle_select();                    return false;                }            }        }                        public function db_close(){            if($this->connection)                mysql_close($this->connection);        }                protected function db_handle_error_connetion(){            echo 'Failed connetion';        }        protected function db_handle_select(){            echo 'Failed access database!';        }    }?>

------
派生类：StudentDetailsDataManager

PHP code

<!--Code highlighting produced by Actipro CodeHighlighter (freeware)http://www.CodeHighlighter.com/-->//StudentDetailsDataManager.php<?php require 'DatabaseManager.php';        class StudentDetailsDataManager extends DatabaseManager{        public function __construct($connection="",$close_flag=true){            parent::__construct($connection, $close_flag);            $this->host = "localhost";            $this->user = "root";            $this->psw = "root";            $this->name = "test";            $this->db_open();            }                public function getStudentInfo($ID,&$data){            //$query = "SELECT * FROM student WHERE ID ='$ID'";            $query = "select * from student where ID = '$ID'";            $result = mysql_query($query);            //print_r($result);            if (!$result) {                echo "result is empty!!";                return false;            }                        $data = mysql_fetch_array($result,MYSQL_ASSOC);            mysql_free_result($result);        }    }?>

----
使用StudentDetailsDataManager实例获取学生信息

PHP code

<!--Code highlighting produced by Actipro CodeHighlighter (freeware)http://www.CodeHighlighter.com/--><?php require_once 'StudentDetailsDataManager.php';    $stuDataManager = new StudentDetailsDataManager();    $ID = "DA123456"; $data=NULL;    $stuDataManager->getStudentInfo($ID, $data);    $stuDataManager->db_close();    echo $data["ID"];?>

------解决方案--------------------
mysql_error看一下就知道了
------解决方案--------------------
public function __construct($connection,$close_flag){
$this->connection = $connection;
$this->connection = $close_flag;
}
这么严重的错误都看不出来？

另外
if (!mysql_select_db($this->name,$this->connection)) {