mysql - 如何检查行是否存在？并为两个表获取它？

Question

我有两个表:votes | 收藏夹。

这是我的结构:

// votes                                    // favorite
+----+---------+---------+-------+          +----+---------+---------+
| id | id_post | id_user | value |          | id | id_post | id_user |
+----+---------+---------+-------+          +----+---------+---------+

当我显示一个帖子时，我需要知道当前用户是否已经对该帖子进行了投票，如果是，它的值是多少(1 或 -1)？我还需要检查这篇文章是否为当前用户所喜爱？那么我该如何实现呢？

我的尝试:

用于检查现有收藏夹:

SELECT EXISTS(SELECT 1 FROM favorites WHERE post_id={} and user_id={} LIMIT 1) as favorite

我也有相同的查询来检查是否有任何投票？但我想知道如何同时检查和获取？我想知道混合两个查询我应该使用 join 吗？

---

编辑:

这里还有posts表的结构:

// posts
+----+-------+---------+----+-------------+-----------------+
| id | title | content | by | total_votes | total_favorites |
+----+-------+---------+----+-------------+-----------------+

Answer 1

与其在已经查询和获取帖子后单独查询以确定当前用户是否投票/收藏，不如通过在查询检索中添加 LEFT JOIN 来显着减少总体查询数量帖子。

SELECT
  `posts`.`id`,
  `title`,
  `content`,
  `by`,
  `total_votes`,
  `total_favorites`,
  -- Use a CASE condition to return a value for votes
  -- based on whether a row is returned by the LEFT JOIN
  -- Change the values like 'has voted' to whatever you want them to be
  CASE 
    WHEN `votes`.`id` IS NOT NULL THEN 'has voted'
    ELSE 'has not voted'
  END AS `user_has_voted`,
  -- Also return the vote value, which will be NULL if no vote exists
  `votes`.`value` AS `vote_value`,
  -- Do the same for favorites
  CASE 
    WHEN `favorites`.`id` IS NOT NULL THEN 'has favorited'
    ELSE 'has not favorited'
  END AS `user_has_favorited`
FROM
  `posts`
  -- Supply the user's id to the ON condition
  -- for both of the joins.
  LEFT JOIN `votes`
    ON `posts`.`id` = `votes`.`id_post`
    AND `votes`.`id_user` = <user_id>
  LEFT JOIN `favorites`
    ON `posts`.`id` = `favorites`.`id_post`
    AND `favorites`.`id_user` = <user_id>

这样做将仅在一个查询中完成所有事情。使用 LEFT JOIN 的想法是，如果存在相关表，它将返回值，否则返回 NULL。因此，在 CASE 语句中，您可以检查 IS NOT NULL 以确定投票/收藏夹是否存在。