语法错误,意外' $ alert' (T_VARIABLE)

时间:2014-05-19 08:59:43

标签: php mysql

嗯,我明白为什么会有这个错误:

Parse error: syntax error, unexpected '$alert' (T_VARIABLE) in C:\wamp\www\envMailAuto\controler.php on line 19

我的控制器:

include 'modele.php';

$email=$_REQUEST["email2"];
if(isset($_REQUEST["check"])) echo $email;
else echo "omged";
$idarticle = 3555;

$DB_host     = "localhost";
   $DB_select   = "envmail";
   $DB_login    = "root";
   $DB_pass     = '';

$connection=mysql_connect($DB_host,$DB_login,$DB_pass); 

$db=mysql_select_db($DB_select, $conn);


alertarticleDAO $alert = new __alertarticle($connection);

我的模特:

Class alertarticleDAO {

var $connection;

public function __alertarticle($mysqlconnection){
$this->connection = $mysqlconnection;
}

public function insert($Idarticle,$email){
$query=" INSERT INTO envmail ( mail_env , id_article , actif )
VALUES ( $Idarticle , $email , 1)";

mysql_query($query,$this->connection);

}


}

2 个答案:

答案 0 :(得分:1)

只需删除无效关键字alertarticleDAO即可。在PHP中,您不会像那样“设置”变量的类型。

但是你的“构造函数”并不完全正确,要么将其命名为类名,要么将其命名为__construct()

因此,您的构造函数应如下所示:

function __construct($mysqlconnection) { ... }

Instanciate it:

$alert = new alertarticleDAO($connection);

答案 1 :(得分:1)

在PHP中,初始化对象时无需指定类型。改变这一行:

alertarticleDAO $alert = new __alertarticle($connection);

进入这个:

$alert = new alertarticle($connection);

并且构造函数定义也是错误的,请更改:

public function __alertarticle($mysqlconnection){

进入这个:

function __construct($mysqlconnection){