时间戳

首页 > 解决方案 > 当我尝试从数据库中提取数据时 PHP 未定义变量

php - 当我尝试从数据库中提取数据时 PHP 未定义变量

问题描述

因此，当我尝试从数据库中提取数据时，我遇到了这些问题。我的数据库连接工作正常。它显示“已建立数据库连接”，并且在我的 index.php 中，我为每个循环编写了一个来获取数据，在我的 HTML 代码中，我将它显示在表中。我得到了这些错误：


Notice: Undefined variable: jokes in C:\xampp\htdocs\comp1321_database\jokes\jokes.html.php on line 16

Warning: Invalid argument supplied for foreach() in C:\xampp\htdocs\comp1321_database\jokes\jokes.html.php on line 16

这是显示数据的 HTML 和 php 代码：

<?php include_once 'admin/includes/helpers.inc.php';?>

<!DOCTYPE html>
<html lang="en">
<head>
     <meta charset="utf-8">
     <title>List of Jokes</title>
</head>
<body>
<p><a href ="/comp1321_database/jokes/form.html.php/">Add your own joke</a></p>     
<p>Here are all the jokes in the database</p>
<!-- into a table -->
        <table border="1">
        <?php foreach ($jokes as $joke): ?>
            <!-- <form action="?deletejoke" method="post">  -->
             <tr>
             <td><?php html($joke['joketext']);?></td>
             <td><?php $display_date = date("D d M Y", strtotime($joke['jokedate']));
             html($display_date); ?>
         </td>
         <td><img height="100px" src="images/<?php html($joke['image']);?>"
            /></td>
            <td><input type="hidden" name="id" value="<?php echo $joke['id'];
            ?>">
            <input type="submit" value="Delete"></td>
        </tr>
    <!-- </form> -->
        <?php endforeach; ?>
        </table>
   <?php include 'admin/includes/footer.inc.html.php';?>   
</body>
</html>

这是 index.php：



<?php
// selection block
include  'admin/includes/db.inc.php';
// 
try
{
    $sql = 'SELECT * FROM joke';
    $result = $pdo->query($sql);

}  catch (PDOException $e) {
    $error = 'Error fetching jokes' . $e->getMessage();
    include 'error.html.php';
    exit();
}
foreach ($result as $row) {
    $jokes[] = array(
        'joketext'=> $row ['joketext'],
        'jokedate'=> $row['joketext'],
        'image'=> $row['image']
    );

}
 include 'jokes.html.php';
?>

非常感谢。

标签： phphtmlmysql

解决方案

在您尝试使用它之前初始化您的笑话变量，即：

$jokes = [];

try
{
    $sql = 'SELECT * FROM joke';
    $result = $pdo->query($sql);

}  catch (PDOException $e) {
    $error = 'Error fetching jokes' . $e->getMessage();
    include 'error.html.php';
    exit();
}
foreach ($result as $row) {
    $jokes[] = array(
        'joketext'=> $row ['joketext'],
        'jokedate'=> $row['joketext'],
        'image'=> $row['image']
    );

}

推荐阅读