如何更改网站源码/软文推广是什么意思
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//DatabaseManager.phpconnection = $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//StudentDetailsDataManager.phphost = "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($result); } }?>
----
使用StudentDetailsDataManager实例获取学生信息
PHP codegetStudentInfo($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)) {
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