PHP MySQLi多查询-命令不同步：MySQLi\u fetch\u数组

我目前正在用PHP创建一个页面，该页面将显示一个用户及其在我的网站高分中的排名

我遇到了一个问题，我的MySQL查询没有按预期运行，但是，@VladBalmos在这里解决了我的问题：

现在我相信我的MySQL查询可以正常运行了，我需要解决PHP代码中出现的问题。这是我的密码：

$userid = (int) $_GET['searched'];
    for ($i = 0; $i < 5; $i++) {
        if ($i == 0) {
            $skill_query = mysqli_query($database, "SELECT uid, overall, overallxp, gamelevel FROM playerstats WHERE uid = ". $userid ." ORDER BY playerstats.overall DESC, playerstats.overallxp DESC") or print(mysqli_error($database));
            $skill_array = mysqli_fetch_array($skill_query);
            //old rank query: $rank_query = mysqli_query($database, "SELECT count(*) + 1 FROM (SELECT uid, overall, overallxp, gamelevel FROM playerstats GROUP BY playerstats.uid) AS x WHERE overall > (SELECT overall FROM playerstats WHERE uid = ". $userid .")") or print(mysqli_error($database));
            $rank_query = mysqli_multi_query($database, "SET @rank=0; SELECT rank, uid, overall, overallxp FROM (SELECT @rank:=@rank + 1 AS rank, uid, overall, overallxp FROM playerstats ORDER BY overall DESC, overallxp DESC) as tmp") or print(mysqli_error($database));
            $rank_array = mysqli_fetch_array($rank_query);
        } else {
            $skill_query = mysqli_query($database, "SELECT uid, ". $skills[$i] .", gamelevel FROM playerstats WHERE uid = ". $userid ." ORDER BY playerstats.". $skills[$i] ." DESC, playerstats.gamelevel DESC") or print(mysqli_error($database));
            $skill_array = mysqli_fetch_array($skill_query);
            $rank_query = mysqli_query($database, "SELECT count(*) + 1 FROM (SELECT uid, ". $skills[$i] .", gamelevel FROM playerstats GROUP BY playerstats.uid) AS x WHERE ". $skills[$i] ." > (SELECT ". $skills[$i] ." FROM playerstats WHERE uid = ". $userid .")") or print(mysqli_error($database));
            $rank_array = mysqli_fetch_array($rank_query);
        }
    
        echo "<tr class='hs_row'>";
        echo "<td align='center'>";
        echo $rank_array[0]; //rank output
        echo "</td>";
        echo "<td align='center'>";
        echo $skill_array[1]; //skilllevel
        echo "</td>";
        echo "<td align='center'>";
        if ($i == 0) {
            echo number_format($skill_array[2]); //skillxp
        } else {
            echo number_format($skill_array[1]); //skillxp
        }
        echo "</td></tr>";
    }

…我收到的错误是：

警告：mysqli_fetch_数组期望参数1是mysqli_结果，布尔值在第156行的[path]/personal.php中给出

命令不同步；现在无法运行此命令

警告：mysqli_fetch_数组期望参数1是mysqli_结果，布尔值在第159行的[path]/personal.php中给出

命令不同步；现在无法运行此命令

警告：mysqli_fetch_数组期望参数1为mysqli_结果，布尔值在第161行的[path]/personal.php中给出

命令不同步；现在无法运行此命令

所以，我的问题是，如何纠正这个错误

$rank_query = mysqli_multi_query($database, $yourQuery);
if($rank_query) {
    while(mysqli_next_result($dabatase)) {
        if($result = mysqli_store_result($database)) {
            while($row = mysqli_fetch_row($result) {
                // do something
            }
        }

    }
}

您必须循环查看多个查询的结果，然后访问每个结果集的行。因为第一个查询只设置了一个变量，所以您必须跳到下一个结果集并获取实际行，这就是为什么我要*whilemysqli\u next\u result*

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阅读

上的文档啊，我想我在尝试时忘记了mysqli_next_结果。谢谢你的帮助！：