MySql：如何从给定条件中获取下一条记录

Question

如果有两张桌子，一张是球员，另一张是球场。

table :: player：

uid:10, name:jack
uid:23, name:david
uid:37, name:james
uid:42, name:nick
uid:58, name:mark

table :: games

gid: 20, last_player: david
gid: 32, last_player: jack
gid: 38, last_player: mark
gid: 43, last_player: nick

如何从给定的gid中找到下一位玩家？ 我正在寻找一些声明：

SELECT p.uid, p.name FROM player p AND onfield o WHERE o.last_player = p.name AND gid=20 ORDER BY uid ASC;

如果gid = 20，答案是uid：37，name：james

如果gid = 38，答案是uid：10，name：jack

问题是，我必须从声明中获取下一条记录，而在“last_player ='mark'”的情况下，我得到第一条记录。

有没有办法在SQL查询中解决这个问题？

Answer 1

如果o.last_player！= mark

   SELECT p.uid, p.name FROM player p, onfield o WHERE o.last_player = p.name 
   AND p.uid > o.guid ORDER BY uid ASC limit 1;

如果o.last_player == mark

  SELECT p.uid, p.name FROM player p, onfield o ORDER BY p.uid ASC limit 1;