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Question

抱歉,如果我应该知道更好,但我已经在这个问题上苦苦挣扎了一段时间.

我有一个300行的mysql数据库.它包含4列,"eventid","player1","player2","得分".在游戏中,player1给予其他人(player2)100分的分数.

我要做的是向登录用户(player1)显示他们已经得分的"player2"表.

我的代码看起来像这样:

$currentuserid = 00001;

$opponent_data = mysql_query("SELECT * FROM `scores` WHERE `player1` = $currentuserid ORDER by score");

$opponent_count = mysql_num_rows($opponent_data);

    echo $opponent_count.'<br>'; // Just to test -> and it shows I have 144 entries in the array, i.e. 144 player 2's that player 1 has scored

$opponent_scores = mysql_fetch_assoc($opponent_data);

$runrows = $opponent_scores;

foreach ($opponent_scores as &$runrows);

    {
    $id = $runrows['eventid'];
    $player2 = $runrows['player2'];
    $score = $runrows['score'];


    echo $player2." got ".$score;

    echo "<br>";

    }

当我运行这一切时,我能看到的是

144

73得到了44

但我希望看到144排"玩家2"得到"玩家2的得分".

我究竟做错了什么？

Answer 1

在for-each循环之后你有一个分号; 不应该在那里.

Answer 2

另外:mysql_fetch_assoc只返回指向结果集中第一行的指针.这就是为什么你最终只打印一行.

将您的代码更改为:

while($opponent=mysql_fetch_assoc($opponent_data)) {
     echo $opponent['player2']." got ".$opponent['score'];
}