PHP/MySqli 检查表 A ID 与表 B ID

Question

我有以下功能，它显示来自数据库的博客文章:

function show_blog_posts() { 
    include('connection.php');
    $sql = "SELECT blog.title, blog.content, blog.posted_by, blog.date, article_comments.comments, article_comments.comment_by
                FROM blog LEFT OUTER JOIN article_comments
                ON blog.content_id = article_comments.comment_id
                WHERE blog.content != ''
                ORDER BY blog.content_id DESC";
    $result = mysqli_query($dbCon, $sql);
    while ($row = mysqli_fetch_array($result)) {
        echo 
            "<h5 class='posted_by'>Posted by " . $posted_by = $row['posted_by'] . " on " . $row['date'] . "</h5>" . 
            "<h1 class='content_headers'>" . $title = $row['title'] . "</h1>" . 
            "<article>" . $content = $row['content'] . "</article>" . 
            "<div class='commented_by'>Posted by: " . $row['comment_by'] . "</div>" . 
            "<div class='comments'>Comments: " . $row['comments'] . "</div>";
    }
}

如您所见，我有两张表，一张是评论，一张是文章。

我希望能够对这些博客文章实现评论功能，我正在使用以下函数将这些评论插入表 B。

function insert_comments($comment_by, $comments) {
    include('db_connection.php');
    $sql =  "INSERT INTO article_comments (comments, comment_by) VALUES ('$comments', '$comment_by') WHERE article_comments.comment_id = blog.content_id";
    mysqli_query($dbCon, $sql);
}

我正在尝试将表 A 的 ID 与表 B 的 ID 链接起来，因此如果 ID 不匹配，用户将无法发帖。如果我与 ID 不匹配，即使没有要发布的文章，用户仍然可以发表评论。

我还编写了以下函数来获取评论表的 ID 和文章表的 ID

function get_article_id($username) { 
    include('db_connection.php');
    $username = sanitize($username);
    $sql = "SELECT content_id FROM `blog` WHERE content_id = '$content_id'";
    $query = mysqli_query($dbCon, $sql);
    return (mysqli_result($query, 0, 'content_id'));
}

function get_comment_id($comment_id) { 
    include('db_connection.php');
    $comment_id = sanitize($username);
    $sql = "SELECT comment_id FROM `article_comments` WHERE comment_id = '$comment_id'";
    $query = mysqli_query($dbCon, $sql);
    return (mysqli_result($query, 0, 'comment_id'));
}

如何检测表 A 的 ID 并将其与表 B 的 ID 进行比较？如果一个

此外，我的代码有什么问题，因为 sql 查询没有执行。

Answer 1

首先我想告诉你一个博客可以有多个评论。因此，您需要仔细处理与评论表的关系。

article_comments 表将有一个外键 ID 作为 "content_id"

现在将您的 show_blog_posts 函数的查询更改为没有评论关系的类似内容，然后使用另一个查询在 while 循环中获取其评论

这是代码。

function show_blog_posts()
{
    include('connection.php');
    $sql = "SELECT blog.title, blog.content, blog.content_id, blog.posted_by, blog.date
            FROM blog
            WHERE blog.content != ''
            ORDER BY blog.content_id DESC";
    $result = mysqli_query($dbCon, $sql);

    if (mysqli_num_rows($result) > 0) {
        while ($row = mysqli_fetch_array($result)) {
            echo
                "<h5 class='posted_by'>Posted by " . $posted_by = $row['posted_by'] . " on " . $row['date'] . "</h5>" .
                "<h1 class='content_headers'>" . $title = $row['title'] . "</h1>" .
                "<article>" . $content = $row['content'] . "</article>";

            $sql = "SELECT article_comments.comments, article_comments.comment_by
            FROM article_comments
            WHERE article_comments.content_id =" . $row['content_id'] . "
            ORDER BY article_comments.comment_id DESC";
            $comments = mysqli_query($dbCon, $sql);

            echo "<div>"; //comments panel
            if (mysqli_num_rows($result) > 0) {
                while ($comment = mysqli_fetch_array($comments)) {
                    echo "<div class='commented_by'>Posted by: " . $comment['comment_by'] . "</div>";
                    echo "<div class='comments'>Comments: " . $comment['comments'] . "</div>";
                }
            }
            echo "</div>"; //comments panel ends
        }
    }

}

要向 article_comments 表插入评论，您还必须在查询中指定 content_id。因此，让您在函数中也传递 content_id。我对您的 insert_comments 函数做了一些更改。插入命令中不需要 where 子句，因为插入命令不支持它

这是代码。

function insert_comments($comment_by, $comments, $content_id) {
    include('db_connection.php');
    $sql =  "INSERT INTO article_comments (content_id, comments, comment_by) VALUES ($content_id, '$comments', '$comment_by');";
    mysqli_query($dbCon, $sql);
}

当您在 get_article_id 中传递 username 时，让您根据 username 而不是 content_id 进行比较。

function get_article_id($username) {
    include('db_connection.php');
    $username = sanitize($username);
    //if username is unique
    $sql = "SELECT content_id FROM `blog` WHERE username = '$username'";
    $query = mysqli_query($dbCon, $sql);
    return (mysqli_result($query, 0, 'content_id'));
}

您必须在 get_comment_id 中传递 username 而不是 comment_id，因为您要检索 comment_id。当您知道 comment_id 时，就不需要调用此函数。我已经相应地改变了它。

function get_comment_id($username) {
    include('db_connection.php');
    $username = sanitize($username);
    $sql = "SELECT comment_id FROM `article_comments` WHERE username = '$username'";
    $query = mysqli_query($dbCon, $sql);
    return (mysqli_result($query, 0, 'comment_id'));
}