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php - 创建排名基础系统并更新排名

转载 作者:行者123 更新时间:2023-11-29 03:02:53 25 4
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伙计们,我是新来的,也是 MySQL 的新手..

所以我正在尝试创建一个管理团队记录的数据库。该数据库包含一个名为 team 的表,其中包含一组列,如下所示,

  • 团队编号
  • 团队排名
  • 团队名称
  • 团队胜利
  • 团队损失
  • 团队积分

所以议程是根据分数对团队进行排名,分数越高排名越高。

<?php
$con = mysqli_connect("", "", "", "");
// Check connection
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

$result = mysqli_query($con, "SET @rownum := 0;
INSERT INTO team( TeamRank, TeamName)
SELECT @rownum := @rownum + 1 AS TeamRank, TeamName
FROM (SELECT SUM(TeamRank)AS TeamRank , TeamName
FROM team
GROUP BY TeamName
ORDER BY TeamRank DESC) as result
ON DUPLICATE KEY UPDATE TeamName = VALUES(TeamName);"
);

echo "<table border='1'>
<tr>
<th>Rank</th>
<th>TeamID</th>
<th>TeamName</th>
<th>Total Points</th>
</tr>";

while ($row = mysqli_fetch_array($result)) {
echo "<tr>";
echo "<td>" . $row['TeamRank'] . "</td>";
echo "<td>" . $row['TeamID'] . "</td>";
echo "<td>" . $row['TeamName'] . "</td>";
echo "<td>" . $row['TeamPoints'] . "</td>";
echo "</tr>";
}

echo "</table>";

mysqli_close($con);

我收到这个错误

Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result

我哪里错了?如果代码有问题,请指导我。

P.S[编辑]:我想做的就是将 TeamName、TeamWins、TeamLoss 和 TeamPoints 作为输入,并且随着团队的 TeamPoints 增加/减少,它应该向上/向下移动排名并显示排名表。

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