PHP連接查詢返回不需要的結果

Question

我的查詢是這樣，php是：

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

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

我知道它不正確的PHP回聲只是更快的寫作

輸出：

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;

該代碼輸出兩個游戲結果，但只想輸入查詢中的玩家，中間插入一個點。

• Games表：ID，游戲，密碼，描述，階梯，區域。
• Players ：id，玩家

我想檢查參加游戲的玩家

Answer 1

我想如果用戶玩更多的游戲，您應該有一個中間表，所以我會有這些表：

games
id
name

players
id
name

player_games
game_id
user_id

我會得到這樣的用戶“ John”的游戲：

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'

可選：在player_games上添加唯一鍵約束