PHP的成員函數致命備調用（）對非目標

Question

到處找一個答案，爲什麼我的PDO準備（）函數是給我這個錯誤：

PHP致命錯誤：調用一個成員函數準備（）在/var/www/database.class.php 26行上的非對象

我一直在尋找關於此問題的每一篇文章，但他們似乎都沒有幫助，甚至只是清除了這個錯誤。我開始在這裏：http://culttt.com/2012/10/01/roll-your-own-pdo-php-class/ 一個還沒有改變任何東西只是試圖用POST數據使用

這是我database.class.php：

<?php 
class Database { 
    private $host = "localhost"; 
    private $user = "nicholas"; 
    private $pass = "12345"; 
    private $dbname = "sstest"; 

    private $dbh; 
    private $error; 
    public $stmt; 

    public function __construct() { 
     $dsn = 'mysql:host=' . $this->host . ';dbname=' . $this->dbname; 
     $options = array(
      PDO::ATTR_PERSISTENT => true, 
      PDO::ATTR_ERRMODE => PDO::ERRMODE_EXCEPTION 
      ); 
     try { 
      $this->dbh = new PDO($dsn, $this->user, $this->pass, $options); 
     } catch (PDOException $e) { 
      $this->error = $e->getMessage();  
     } 
    } 

    public function query($query) { 
     $this->stmt = $this->dbh->prepare($query); //this is line 26 
    } 

    public function bind($param, $value, $type = null) { 
     if (is_null($type)) { 
      switch (true) { 
      case is_int($value): 
       $type = PDO::PARAM_INT; 
       break; 
      case is_bool($value): 
       $type = PDO::PARAM_NULL; 
       break; 
      default: 
       $type = PDO::PARAM_STR; 
       break; 
      } 
     } 
     $this->stmt->bindValue($param, $value, $type); 
    } 

    public function execute() { 
     return $this->stmt->execute(); 
    } 

    public function resultSet() { 
     $this->execute(); 
     return $this->stmt->fetchAll(PDO::FETCH_ASSOC); 
    } 

    public function single() { 
     $this->execute(); 
     return $this->stmt->fetch(PDO_::FETCH_ASSOC); 
    } 

    public function rowCount() { 
     return $this->stmt->rowCount(); 
    } 

    public function lastInsertId() { 
     return $this->dbh->lastInsertId(); 
    } 

    public function beginTransaction() { 
     return $this->dbh->beginTransaction(); 
    } 

    public function endTransaction() { 
     return $this->dbh->commit(); 
    } 

    public function cancelTransaction() { 
     return $this->dbh->rollBack(); 
    } 


} 

?> 

，這裏是用我的數據庫中的PHP文件：

<?php 
include 'database.class.php'; 

$id = $_POST["id"]; 
$fname = $_POST["fname"]; 
$lname = $_POST["lname"]; 

$database = new Database(); 

$sql = 'INSERT INTO sstest (id, fname, lname) VALUES (:id, :fname, :lname)'; 

$database->query($sql); 

$database->bind(':id', $id); 
$database->bind(':fname', $fname); 
$database->bind(':lname', $lname); 

$database->execute(); 

echo $database->lastInsertId(); 
?> 

我知道java代碼發送POST數據，但我很困惑爲什麼prepare（）函數說$ sql是一個非對象。任何幫助非常感謝，我一直在爲此工作2天，並且無法超越prepare（）語句。

Answer 1

我覺得你的問題是在這裏：

try { 
    $this->dbh = new PDO($dsn, $this->user, $this->pass, $options); 
} catch (PDOException $e) { 
    $this->error = $e->getMessage();  
} 

加入這一行的行後趕上你那裏：

var_dump($this->error);exit();