致命错误：在null上调用成员函数prepare（）

Question

我正在尝试访问类别及其内容的列表。我有一堂课叫做分类。我不断收到此错误。奇怪的是，到目前为止，我已经在另外两个地方使用了相同的完全相同的代码，没有任何问题。我在这里所做的只是重用代码并更改所有变量。

Fatal error: Call to a member function prepare() on null

这是我班上的代码：

    <?php

class Category {

    public function fetch_all() {

        global $pdo;

        $query = $pdo->prepare("SELECT * FROM dd_cat");

        $query->execute();

        return $query->fetchAll();

    }

    public function fetch_data($cat_id) {

        global $pdo;

        $query = $pdo->prepare("SELECT * FROM dd_cat WHERE cat_id = ?");

        $query->bindValue(1, $cat_id);

        $query->execute();

        return $query->fetch();

    }

}

?>

这是我尝试调用的代码：

<?php

session_start();

//Add session_start to top of each page//

require_once('includes/config.php');

require_once('includes/header.php');

include_once('includes/category.php');

?>

<link rel="stylesheet" href="css/dd.css">

    <div id="menu">

        <a class="item" href="drop_index.php">Home</a> -

        <a class="item" href="create_topic.php">Create a topic</a> -

        <a class="item" href="create_cat.php">Create a category</a>

        <div id="userbar">

<?php

    if( $user->is_logged_in() )

    {

        echo 'Hello ' . $_SESSION['user_name'] . '. Not you? <a href="logout.php">Sign out</a>';

    }

    else

    {

        echo '<a href="login.php">Sign in</a> or <a href="index.php">create an account</a>.';

    } 

?>

        </div>

    </div>

<?php

$category = new Category;

$categories = $category->fetch_all();

?>

    <div id ="wrapper">

        <h1>Categories</h1>

        <section>

            <ul>

                <?php foreach ($categories as $category) { ?>

                    <li><a href="category.php?id=<?php echo $category['cat_id']; ?>">

                        <?php echo $category['cat_title']; ?></a> 

                    </li>

                <?php } ?>

            </ul>

        </section>

    </div>

<?php

require_once('includes/footer.php');

?>

Answer 1

看来您的$pdo变量未初始化。在您上传的代码中，我看不到它的初始化位置。确保global scope在调用类方法之前在中创建了一个新的PDO对象。（由于应在Category类中实现方法，因此应在全局范围内声明它）。$pdo = new PDO('mysql:host=localhost;dbname=test', $user, $pass);

Answer 2

您可以尝试/捕获PDOException（您的配置可能有所不同，但重要的部分是尝试/捕获）：try {&nbsp; &nbsp; &nbsp; &nbsp; $dbh = new PDO(&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; DB_TYPE . ':host=' . DB_HOST . ';dbname=' . DB_NAME . ';charset=' . DB_CHARSET,&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; DB_USER,&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; DB_PASS,&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; [&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; PDO::ATTR_PERSISTENT&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; => true,&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; PDO::ATTR_ERRMODE&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;=> PDO::ERRMODE_EXCEPTION,&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; PDO::MYSQL_ATTR_INIT_COMMAND&nbsp; &nbsp; => 'SET NAMES ' . DB_CHARSET . ' COLLATE ' . DB_COLLATE&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; ]&nbsp; &nbsp; &nbsp; &nbsp; );&nbsp; &nbsp; } catch ( PDOException $e ) {&nbsp; &nbsp; &nbsp; &nbsp; echo 'ERROR!';&nbsp; &nbsp; &nbsp; &nbsp; print_r( $e );&nbsp; &nbsp; }该print_r( $e );行将向您显示所需的一切，例如，我最近遇到了错误消息如的情况unknown database 'my_db'。