mysql_query() 執(zhí)行結(jié)果一直為false

1)現(xiàn)有一個數(shù)據(jù)庫名為test，里面只有一個表student。

屬性名稱：ID, Name, Email.

2)嘗試著將數(shù)據(jù)庫連接與操作封裝成一個類DatabaseManager,并擴展了一個類StudentDetailsDataManager來獲取學生信息。

3)問題：能夠連接到test數(shù)據(jù)庫，sql語句在數(shù)據(jù)庫中測試過沒有問題，但mysql_query()執(zhí)行sql語句結(jié)果一直為false。不知什么問題？

代碼如下：

數(shù)據(jù)庫操作基類：DatabaseManager

PHP code//DatabaseManager.php

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!';

}

}

?>

------

派生類：StudentDetailsDataManager

PHP code//StudentDetailsDataManager.php

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實例獲取學生信息

PHP codegetStudentInfo($ID, $data);

$stuDataManager->db_close();

echo $data["ID"];

?>

------解決方案--------------------

mysql_error看一下就知道了

------解決方案--------------------

public function __construct($connection,$close_flag){

$this->connection = $connection;

$this->connection = $close_flag;

}

這么嚴重的錯誤都看不出來？

另外

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

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