php - PHP/MySqli根据表B ID检查表ID

我有以下功能，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>";
    }
}

如你所见，我有两张桌子，一张有评论，一张有文章。
我想能够实现对这些博客文章的评论功能，我使用下面的函数将这些评论插入到表B中。

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 ID与表B ID链接，因此如果ID不匹配，用户将无法发布。如果我与ID不匹配，即使没有文章可以发表，用户仍然可以发表评论。
我还编写了以下函数来获取comments表的ID和articles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 id并将其与表B id进行比较？如果
另外，我的代码有什么问题，因为sql查询没有执行。

最佳答案

首先我想告诉你一个博客可以有多个评论。因此，您需要仔细地与注释表建立关系。
article_comments表的外键id将为"content_id"
现在将show_blog_posts函数的查询更改为类似这样的无注释关系的查询，然后使用另一个查询在while循环中获取其注释
这是密码。

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表插入注释，还必须在查询中指定content\u id。所以，让您也在函数中传递内容id。我对您的insert_comments功能做了一些更改。在insert命令中不需要where子句，因为在insert命令中不支持它
这是密码。

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);
}

当您将username传入get_article_id时，让您根据username而不是与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'));
}

您必须传递username而不是comment_id中的get_comment_id，因为您将检索comment_id。当您知道comment_id时，就不需要调用此函数。我已经相应地改了。

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'));
}

关于php - PHP/MySqli根据表B ID检查表ID，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/32018495/