我是 php 新手并试图添加一个删除按钮以从列表中删除一个对象(作业),我希望删除按钮出现在每个单独的对象(作业)旁边,一旦单击此作业就会从数据库中删除桌子。下面是我的edit_jobs.php(显示特定用户的所有作业)和delete_job.php(假设从表中删除该特定作业)的代码有人可以告诉我我做错了什么,
我的 edit_jobs 页面显示特定用户发布的表中的所有工作。
<?php
        include_once "connect_to_mysql.php";
        $id = $userid;
        $username = $_GET['username'];
        $result = mysql_query("SELECT * FROM jobs WHERE user_id ='$id'")
                or die(mysql_error());
        while ($row = mysql_fetch_array($result)) {
            echo '<a href="job.php?id=' . $row['job_id'] . '"> ' . $row['job'] . '</a><br />';
            echo 'category: ' . $row['category'] . '<br />';
            echo 'description: ' . $row['description'] . '<br />';
            echo '<a href="member.php?id=' . $row['userid'] . '">Clients profile</a><br />';
            echo '<br />';?>
        <a href="delete_job.php?job_id=<?php echo $row['job']; ?>"
                       onclick="return confirm('Are you sure you want to delete this book?');">
                       <img src="images/delete20.png" alt="Delete Book" />
                    </a>
        <?php } ?>
delete_job 页面
<?php
if ($_SERVER['REQUEST_METHOD'] == 'GET') {
if (!empty($_GET['job_id'])) {
    $jobId = $_GET['job_id'];
    require_once 'connect_to_mysql.php';
    $sql = "DELETE FROM jobs WHERE job_id = ?";
    $params = array($jobId);
    $stmt = $link->prepare($sql);
    $status = $stmt->execute($params);
    if ($status == true) {
        header("Location: edit_jobs.php");
    }
    else {
        $error_info = $stmt->errorInfo();
        $error_message = "failed to delete job: {$error_info[2]} - error code {$error_info[0]}";
        require 'error.php';
    }
}
else {
    $error_message = "book id not specified";
    require 'edit_jobs.php';
}
}
else {
}
?>