Mysql 获取当前行以及下一个和上一个记录

我有一个查询，它将获取当前记录以及下一个和上一个记录。但我需要这样做，这样它就不必是下一个连续的ID

select u.user_id, u.first_name, u.last_name, u.username,
    (select user_id -1 as user_id from users where user_id = u.user_id) as prev_id,
    (select user_id +1 as user_id from users where user_id = u.user_id) as next_id
from users u
where u.user_id = 3

这将返回我需要的内容，但如果用户_id 4不存在，我将如何获取下一条记录

还有，这是最快的方法吗

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感谢您的帮助？

您可以使用limit获取下一个和上一个id。例如：

  select
          u.user_id,
          u.first_name,
          u.last_name,
          u.username,
          (select user_id as user_id 
           from users 
           where user_id <u.user_id 
           order by user_id desc
           limit 1 ) as prev_id,
           (select user_id as user_id 
           from users 
           where user_id >u.user_id 
           order by user_id asc
           limit 1 ) as next_id
        from users u
        where u.user_id = 3

---

使用两个联接和聚合函数：-

select u.user_id, u.first_name, u.last_name, u.username,
    MAX(u_prev.user_id) AS prev_id,
    MIN(u_next.user_id) AS next_id
from users u
LEFT OUTER JOIN user u_prev ON u.user_id > u_prev.user_id
LEFT OUTER JOIN user u_next ON u.user_id < u_next.user_id
where u.user_id = 3
GROUP BY u.user_id, u.first_name, u.last_name, u.username

---

希望这能对您有所帮助。

在应用程序中使用数据比查询本身复杂多了

SELECT * 
FROM `user` 
WHERE id >= (
 SELECT id 
 FROM `user` 
 WHERE id < (SELECT id FROM `user` WHERE firstname = 'abc')
 ORDER BY id DESC 
 LIMIT 1
)
ORDER BY id ASC
LIMIT 3