-2

我试图在数据库中插入具有唯一 ID 的播放器。这个模型对我有用，但如果我给它一个密钥更新，它会在我的数据库中复制它。

选项1

$id = uniqid();  
$insert_player_query    = mysqli_query($db,"INSERT INTO players (id,nickname,score,time_online,mapname,sid) VALUES ('$id','$player_nickname','$player_score','$player_time','$mapname','$server_id') ON DUPLICATE KEY UPDATE score = score + VALUES(score), time_online = time_online + VALUES(time_online) WHERE id='$id'");

选项 2

然后我尝试了，但它根本没有插入任何东西

$id = uniqid();  
$sql=mysqli_query($db, "SELECT * FROM players WHERE nickname='$player_nickname'");

if (mysqli_num_rows($sql) > 0) 
    {
         $insert_player_query2  = mysqli_query($db,"UPDATE players (nickname,score,time_online,mapname,sid) VALUES ($player_nickname','$player_score','$player_time','$mapname','$server_id') WHERE id='$id'");   
                
    } else {
         $insert_player_query   = mysqli_query($db,"INSERT INTO players (id,nickname,score,time_online,mapname,sid) VALUES ('$id','$player_nickname','$player_score','$player_time','$mapname','$server_id') WHERE id='$id'");   
    }
4

2 回答 2

0

删除“where”条件。它在“INSERT”语句中没有任何意义。

于 2020-07-20T10:06:44.860 回答
0

您应该检查字段并使用：

CREATE TABLE `players` (
  `id` int(11) unsigned NOT NULL AUTO_INCREMENT,
  `nickname` varchar(100) DEFAULT NULL,
  `score` int(11) DEFAULT NULL,
  `time_online` varchar(100) DEFAULT NULL,
  `mapname` varchar(100) DEFAULT NULL,
  `sid` int(11) DEFAULT NULL,
  PRIMARY KEY (`id`),
  UNIQUE KEY `uniq` (`id`)
) ENGINE=InnoDB AUTO_INCREMENT=2 DEFAULT CHARSET=utf8;


INSERT INTO players (id,nickname,score,time_online,mapname,sid) VALUES ('1','player_nickname','1','1','mapname','1') 
ON DUPLICATE KEY UPDATE score = score + VALUES(score), time_online = time_online + VALUES(time_online)

请尝试一下。

于 2020-07-20T09:31:52.403 回答