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列表.我有一个名为Categories的课程.我一直收到这个错误.奇怪的是,到目前为止,我已经在其他两个地方使用了相同的精确代码而没有任何问题.我在这里所做的就是重用代码并更改所有变量.

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

这是我班级的代码:

    <?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();
    }
}

?>

这是我要调用的代码:

<?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

看起来您的$pdo变量未初始化.我在你初始化它的代码中看不到.

确保global scope在调用类方法之前创建新的PDO对象.(您应该在全局范围内声明它,因为您在Category类中实现了这些方法).

$pdo = new PDO('mysql:host=localhost;dbname=test', $user, $pass);