Question

我遵循了关于PHP的书中的代码，但我遇到了以下问题：它显示了我的数据库中的帖子，但是当我点击＆＃34;阅读更多＆＃34;链接它告诉我它无法找到URL。以下是代码

任何有关错误的建议都会非常感激。

posts.php

<?php
Class Posts {
    public $db = '';
    public function __construct(){

        $this->db = new PDO("mysql:host=localhost;dbname=testposts", "username", "password");
        $this->db->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
        $this->index(); 

    }

    public function index(){
        $id = 0;
        $posts = array();
        $template = '';
        if (!empty($_GET['id'])){
            $id = $_GET['id'];
        }

        try {
            if (!empty($id)){
                $query = $this->db->prepare("SELECT * FROM posts WHERE id = ?");
                $params = array($id);
                $template = 'single-post.php';
            } else {
                $query = $this->db->prepare("SELECT * FROM posts");
                $params = array();
                $template = 'list-posts.php';
            }
            $query->execute($params);
            for ($i = 0; $row = $query->fetch(); $i++) {
                $posts[] = array('id' => $row['id'], 'content' => $row['content']);
            }
        } catch (PDOException $e){
            $e->getMessage();
        }
        $query->closeCursor();
        $db = null;
        require_once($template);
    }
}

$posts = new Posts();

?>

列表posts.php

 <h1>List of Blog Posts</h1>
    <?php foreach ($posts as $post): ?>
        <h3>Post # <?php echo htmlspecialchars($post['id']); ?></h3>
        <?php echo htmlspecialchars($post['content']); ?>
        <a href="http://localhost/other/posts.php?id=<?php 
            echo htmlspecialchars($post['id']); ?>">Read More</a>
        <hr>
    <?php endforeach; ?>

单post.php中

<?php foreach ($posts as $post): ?>
    <h1>Post #<?php echo htmlspecialchars($post['id']); ?></h1>
    <hr>
    <?php echo htmlspecialchars($post['content']); ?>
    <a href="http://localhost/other/posts.php">Back to Post List</a>
<?php endforeach; ?>

PHP PDO - 如何解决此问题？

0 个答案: