连接数据库失败


Connection to database failure

我似乎无法弄清楚我的数据库或连接到它的问题。当我试图运行它时,我得到了这个错误。

注意:试图获得非对象的属性在/Applications/XAMPP/xamppfiles/htdocs/swiftappandmysql/db/mysqlda.php行54

致命错误:未捕获异常' exception '在/Applications/XAMPP/xamppfiles/htdocs/swiftappandmysql/db/MySQLDAO.php:53堆栈跟踪:#0/Applications/XAMPP/xamppfiles/htdocs/swiftappandmysql/scripts/registerUser.php(41): MySQLDAO->registerUser('email', 'aske', 'meyer', '1bf528e7f15d11c…', 'KO'x8E'xD0'xCE/'xBD'xACK'xD1d'x18'x9A'x07'xE1…')#1 {main}抛出在/Applications/XAMPP/xamppfiles/htdocs/swiftappandmysql/db/mysqlda1 .php第53行

连接文件:

<?php class Conn { public static $dbhost = "localhost";
        public static $dbuser = "root";
        public static $dbpass = "";
        public static $dbname = "app";} ?>

MySQL文件

public function registerUser($email, $first_name, $last_name, $password, $salt)
{ 
    $sql = "insert into users set email=?, first_name=?, last_name=?, user_password=?, salt=?";
    $statement = $this->conn->prepare($sql);
    if (!$statement)
      throw new Exception($statement->error);
    $statement->bind_param("sssss", $email, $first_name, $last_name, $password, $salt);
    $returnValue = $statement->execute();
    return $returnValue;  
}   

第54行。if (!$statement) throw new Exception($statement->error);

if()块在$statement = FALSE时运行,因此$statement->error不可能正确。error属性在$this->conn中,所以它应该是

if (!$statement) {
    throw new Exception($this->conn->error);
}

错误信息将显示SQL的问题