Question

我有3张桌子：

users
  user_id
  first_name
forum_post
 post_id
 post_title
user_post_join
  id
  user_id
  post_id

连接表从用户获取user_id，从forum_posts获取post_id。

我想要显示的是forum_post表和发布它的用户的帖子列表。我很难知道加入应该在哪里，这是我到目前为止的尝试......

function build_forum_posts(){

        global $dbc;

        $q = "SELECT users.user_id, forum_post.post_id, user_post_join.id 
            FROM users 
            INNER JOIN user_post_join ON users.user_id = forum_posts.post_id
             ";  

        $r = mysqli_query ($dbc, $q); // Run the query.

        // FETCH AND PRINT ALL THE RECORDS
        while ($row = mysqli_fetch_array($r)) {

        echo '
        <div class="post">
            <div class="col-group-2">
                <h3>'.$row["post_id"]. '</h3>
                <p>By: '.$row["user_id"]. '</p>
            </div>
            <div class="col-group-2">
                <div class="post_count">
                    <h3  class="answer">0</h3>
                    <p class="answer">Answers</p>
                </div>
            </div>  
        </div>
        ';

        }

Answer 1

您没有加入所有三张桌子。尝试这样的事情：

SELECT u.user_id, u.first_name, f.post_id, f.post_title, j.id 
 FROM users AS u
 INNER JOIN user_post_join AS j ON u.user_id = j.user_id
 INNER JOIN forum_posts AS f ON f.post_id = j.post_id

从表连接中选择

1 个答案: