Php 排列联接数据库结果

我目前正在从我的“social_posts”表中获取详细信息，然后将其标签、喜欢的数量以及答案的数量添加到结果对象中。有没有一种方法可以做到这一点，而不必在循环中进行额外的查询

    $query = "SELECT * FROM social_posts JOIN users ON social_posts.user_id = users.id";
    $posts = $this->db->query($query);

    if ($posts->num_rows() > 0) {        
        foreach ($posts->result() as $p => $post) {                     
            // Get the question's tags
            $tags = $this->db->query("SELECT * FROM social_tags 
                WHERE post_id = ?", $post->post_id);

            // Get the number of likes
            $likes = $this->db->query("SELECT id FROM social_likes 
                WHERE post_id = ?", $post->post_id);

            // Get the number of answers
            $answers = $this->db->query("SELECT id FROM social_responses 
                WHERE post_id = ?", $post->post_id);

            $post->tags = $tags->result();
            $post->likes = $likes->num_rows();
            $post->answers = $answers->num_rows();
            $post->author = array(
                "firstname" => $post->firstname,
                "thumbnail" => $post->thumbnail,
            );
        }

        return $posts->result();
    } else {
        return FALSE;
    }

您可以尝试以下SQL：

SELECT 
    social_posts.*,
    users.*,
    GROUP_CONCAT(social_tags.name) AS tags,
    COUNT(social_likes.id) AS likes,
    COUNT(social_responses.id) AS answers
FROM 
    social_posts 
        JOIN users ON social_posts.user_id = users.id
        LEFT JOIN social_tags ON social_tags.post_id = social_posts.id
        LEFT JOIN social_likes ON social_likes.post_id = social_posts.id
        LEFT JOIN social_responses ON social_responses.post_id = social_posts.id
GROUP BY
    social_posts.id

---

您将获得逗号分隔字符串形式的标记。当然，您需要调整列名以适应您的数据库。

非常感谢，这太棒了！但它似乎没有显示任何对帖子的回复？我应该在哪里指定“where”子句呢？1）在您的原始代码中，据我所知，您只得到了响应的数量。可能更好的办法是进行后续查询，以获取给定帖子的所有回复，因为您可能希望从social_responses表中获得更多列。2）WHERE子句介于from和GROUP BY之间。您的说法是正确的，我只需要回复的计数，对不起。我在应用程序中看到了一个不同的函数，弄糊涂了。我想我已经把它整理好了。谢谢！组_CONCAT（DISTINCT social_tags.name）应该可以工作，请查看