PHP 连接查询返回不需要的结果

Question

我的查询是这样的，php 是:

Select * 
from games 
left join players on games.id = players.id

echo "game:" .  $player . ",pass" . $pass...

我知道它不正确 php echo 只是更快的写入

输出:

game:ost1232,pass:10,desc:Difficulty:Highest,players:Duiski,lad:0region:3;game:ost1232,pass:10,desc:Difficulty:Highest,players:testarn,lad:0,region:3;

但我想要

game:ost1232,pass:10,desc:Difficulty:Highest,players:Duiski.testarn,lad:0,region:3;

代码输出两个游戏结果，但只想在查询中输入玩家，中间用点表示。

• 游戏表:id、游戏、密码、描述、天梯、地区。
• 玩家:id、玩家

我想查看加入游戏的玩家

Answer 1

我猜如果用户玩更多游戏，你应该有一个中间表，所以我会有这些表:

games
id
name

players
id
name

player_games
game_id
user_id

我会像这样获取用户“John”的游戏:

select 
players.name as player_name,
games.name as game_name
from games 
join player_games on player_games.game_id
join players on player_games.user_id = players.id
where players.name='John'

可选:在player_games上添加唯一键约束