Question

我正在尝试从当前用户关注的人员及其详细信息中检索帖子。我一直在研究，似乎加入是正确的方法，但我不确定它有什么问题。

这是我的代码

$queryfeed = "SELECT posts.user_id, posts.body, posts.image, posts.has_image 
FROM posts, follows WHERE posts.user_id =  follows.leader_id 
AND follows.follower_id =$user_id INNER JOIN users ON users.user_id 
= follows.leader_id";

$resultfeed = $db->query($queryfeed);
if($resultfeed->num_rows > 0) {
    while( $rowfeed = $resultfeed->fetch_assoc() ) {       
        if($rowfeed['posts.has_image'] == 1) {
    ?>
            <article class="post">
                <div class="post-head cf">
                    <a class="userpic" href=""><img src="<?php echo $userpic ?>" alt="<?php echo $rowfeed['users.username'] ?>"></a>
                    <a href="" class="username">
                        <?php echo $rowfeed[users.'username']; ?>
                    </a>
                </div>
                <img src="users/user_<?php echo $rowfeed['posts.user_id'] ?>/posts/<?php echo $rowfeed['posts.image']; ?>" alt="">
                <div class="post-body">
                    <div class="post-options">
                        <a class="likes" href="">2 likes</a>
                    </div>
                    <p>
                        <a class="username" href="">
                                <?php echo $rowfeed['users.username'] ?>
                        </a>
                        <?php echo $rowfeed['posts.body'] ?>
                    </p>
                    <hr />
                    <div class="cf">
                        <a class="like hide-text" href="javascript:;">Like This Post</a>
                        <form action="" class="comment">
                            <input type="text" placeholder="Add a comment">
                        </form>
                    </div>
                </div>
            </article>

   <?php } else { ?>

我的表格看起来像这样

users
user_id | username | avatar

posts
post_id | user_id | body | image | has_image

follows
leader_id | follower_id

Answer 1

从您的原始查询中，我发现您只对帖子的数据感兴趣（而不是发布帖子的用户的数据）。我假设$ user_id保存当前用户的id，即也用作follower_id的值。如果是这种情况，查询会变得更容易：

SELECT p.user_id, p.body, p.image, p.has_image
FROM followers f INNER JOIN posts p ON f.leader_id = p.user_id and f.follower_id = $user_id

如果您对海报的用户数据感兴趣，请按以下步骤更改查询：

SELECT p.user_id, p.body, p.image, p.has_image, u.username, u.avatar
FROM follows f
  INNER JOIN posts p ON f.leader_id = p.user_id and f.follower_id = $user_id
  INNER JOIN users u ON p.user_id = u.user_id

Answer 2

我现在无法访问计算机，但似乎您的SQL-Query有一些错误。看看我的更正版本：

SELECT posts.user_id, posts.body, posts.image, posts.has_image FROM posts
INNER JOIN follows ON posts.user_id = follows.follower_id
WHERE follows.leader_id=$user_id;

Answer 3

尝试此查询：

SELECT
    posts.user_id, posts.body, posts.image, posts.has_image
    authors.avatar, authors.username
FROM
    posts
LEFT JOIN
    users as authors
ON
    authors.user_id=posts.user_id
LEFT JOIN
    follows
ON
    follows.leader_id=authors.user_id
WHERE
    follows.follower_id={$user_id}

仅使用LEFT JOIN ....

Answer 4

让我们假设当前的user.user_id为1，关注者为user.user_id 2,3和4

我认为我们需要符合以下标准吗？

posts.user_id = follows.follower_id follows.leader_id = $user_id

所以查询需要这样吗？

SELECT * FROM posts INNER JOIN follows ON posts.user_id = follows.follower_id INNER JOIN users ON follows.leader_id = $user_id

以下是表格的模拟示例

如何将内容加入用户表跟随用户帖子？

4 个答案: