识别和连接表

Question

之间ID的I有两个表 - 一个被存储文章内容，另一种是存储文章评论

我使用来显示这些功能是：

function list_articles() { 
    include('core/db/db_connection.php'); 
    $sql = "SELECT blog.content_id, 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.blog_id 
      WHERE blog.content != '' 
      ORDER BY blog.content_id DESC"; 
    $result = mysqli_query($dbCon, $sql); 

    $previous_blog_id = 0; 

    while ($row = mysqli_fetch_array($result)) { 
     if ($previous_blog_id != $row['content_id']) { 
      echo "<h5 class='posted_by'>Posted by {$row['posted_by']} on {$row['date']}</h5> 
       <h1 class='content_headers'>{$row['title']}</h1> 
       <article>{$row['content']}</article> 
       <hr class='artline'>"; 
      $previous_blog_id = $row['content_id']; 
     } 
     if (!empty($row['comment_by']) && !empty($row['comments'])) { 
      echo "<div class='commented_by'>Posted by: {$row['comment_by']} </div> 
        <div class='comments'>Comments: {$row['comments']}</div> 
        <hr class='artline2'>"; 
     } 
    } 
} 

我使用下面插入注释article_comments表：

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

这工作 - 它不插入，但我有我怎么能瞄准$ blog_id变量没有线索时，用户提交后...以下是我的形式使用

<?php echo list_articles(); 
    if (!empty($_POST)) { 
     insert_comments($_POST['comments'], $_POST['username'], 11); 
     } 
?> 
<form method='post' action='' class='comments_form'> 
    <input type='text' name='username' placeholder='your name... *' id='name'> 
    <textarea name='comments' id='textarea' placeholder='your comment... *' cols='30' rows='6'></textarea> 
    <input type='submit' name='submit' id='post' value='post'> 
</form> 

我敢打赌，你注意到了我手动插入11作为最后一个变量设置了一个param。这链接到我的article_comments表中的blog_id 11（外键）。它显示评论就好。

有没有办法以$ blog_id为目标，而无需手动插入数字？类似于我如何使用$_POST['comments']来定位$ comments变量？

此外，即使我可以瞄准，我怎么知道哪些帖子是用户评论？我应该给他们在下拉列表中选择的选项吗？这似乎很尴尬..但它是我能想到的唯一解决方案。

编辑：我在一个隐藏字段瞄准blog_id尝试：

function list_articles() { 
    include('core/db/db_connection.php'); 
    $sql = "SELECT blog.content_id, 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.blog_id 
      WHERE blog.content != '' 
      ORDER BY blog.content_id DESC"; 
    $result = mysqli_query($dbCon, $sql); 

    $previous_blog_id = 0; 

    while ($row = mysqli_fetch_array($result)) { 
     if ($previous_blog_id != $row['content_id']) { 
      echo "<h5 class='posted_by'>Posted by {$row['posted_by']} on {$row['date']}</h5> 
       <h1 class='content_headers'>{$row['title']}</h1> 
       <article>{$row['content']}</article> 
       <hr class='artline'>"; 
      $previous_blog_id = $row['content_id']; 
     } 
     if (!empty($row['comment_by']) && !empty($row['comments'])) { 
      echo "<div class='commented_by'>Posted by: {$row['comment_by']} </div> 
        <div class='comments'>Comment: {$row['comments']}</div> 
        <hr class='artline2'>"; 
     } 
     $sql2 = "SELECT FROM article_comments VALUES blog_id"; 
     $result2 = mysqli_query($dbCon, $sql2); 
     while ($row = mysqli_fetch_assoc($result2)) { 
      echo " <form method='post' action='' class='comments_form'> 
         <input type='text' name='username' placeholder='your name... *' id='name'> 
         <textarea name='comments' id='textarea' placeholder='your comment... *' cols='30' rows='6'></textarea> 
         <input type='hidden' name=blog_id' value='{$row['blog_id']}'> 
         <input type='submit' name='submit' id='post' value='post'> 
        </form>"; 
     }     
    } 
} 

SQL2和RESULT2部分是语句导致该错误的

编辑2：

我不认为$ sql2是正确的方法。现在的代码工作正常，但我回到了方块1.对于每个评论插入的文章得到重复。

<form method='post' action='' class='comments_form'> 
    <input type='text' name='username' placeholder='your name... *' id='name'> 
    <textarea name='comments' id='textarea' placeholder='your comment... *' cols='30' rows='6'></textarea> 
    <input type='hidden' name=blog_id' value='{$row['blog_id']}'> 
    <input type='submit' name='submit' id='post' value='post'> 
</form>"; 

有没有办法为目标，而无需调用而（$行= mysqli_fetch_array（$结果））{}的blog_id？或者至少，不在第二个while循环中调用它？

随着第一段代码我贴我得到如下结果：

Article title: LOREM IPSUM 
Content: LOREM IPSUM DOLOR SIT AMET.... 
-------------------------------------- 
Name: DSK 
Comment: Great article! 
-------------------------------------- 
Name: DSK 
Comment: Great article! - 2nd comment 

-- BEGIN SECOND ARTICLE ON WEBPAGE 

Article title: LOREM IPSUM 2nd article 
Content: LOREM IPSUM DOLOR SIT AMET.... 
-------------------------------------- 
Name: User0 
Comment: Great article! 
-------------------------------------- 
Name: User1 
Comment: Great article! - 2nd comment 
-------------------------------------- 
Name: User2 
Comment: Great article! - 3rd comment 
-------------------------------------- 

这正是我要找的。不过，我只能通过phpmyadmin接口插入注释，手动选择外键（blog_id）。

我希望能够通过一个形式得到相同的结果：

Article title: LOREM IPSUM 
Content: LOREM IPSUM DOLOR SIT AMET.... 
-------------------------------------- //comments 
Name: DSK 
Comment: Great article! 
-------------------------------------- 
Name: DSK 
Comment: Great article! - 2nd comment 
-------------------------------------- // end comments 

|-------------------------------------| // comments form 
|Name: New User      | 
|Comment: New comment !    | 
|          | 
|-------------------------------------| 
[Submit] 

当用户提交表单时，他的名字和他的评论被提交到数据库到article_comments表。外键（blog_id）也应链接到现有文章（它的作用）。我只需要一种方法将其定位在我的功能中。

这是否有意义？....

Answer 1

也许你可以使用一个隐藏的表单元素表单里面：

<input type="hidden" name="blog_id" value="<?PHP echo $id;?>"> 

然后在提交你可以用$ _ POST访问[“blog_id”]

如果我理解正确你的问题是。