Question

我正在尝试从数据库的选定表中填充json数组。主表包含用户的所有信息，我想从中选择10个具有顶级XP的成员。

我的代码：

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    clicked_count = 0;
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        cur_anchor = anchors[i];
        cur_anchor.setAttribute("class", i);
        cur_anchor.onclick = function (e) {
            e = e || window.event;
            e.preventDefault();

            my_anchor_id = this.getAttribute("class");
            if (!(my_anchor_id in clicked)) {
                clicked[my_anchor_id] = null;
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                }
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        };
    }
};

Answer 1

首先，没有理由检查是否发布了USERNAME，因为您从未使用过USERNAME字段。其次，你错过了一个收尾卷曲。

至少，你应该重新排序你的sql语句 - LIMIT之后使用ORDER BY

$serverName = "MyServerName";
$dbUsername = "DatabaseUsername";
$dbPassword = "DatabasePassword";
$dbName = "DatabaseID";
$tableName = "TableName";

$conn = new mysqli($serverName, $dbUsername, $dbPassword, $dbName);
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}
try {
    $sqlTopMembers = "SELECT XP, username FROM $tableName ORDER BY XP DESC LIMIT 10";
    if (!($result = $conn->query($sql)))
        die($conn->error);
    if ($result->num_rows) {
        $rankings = [];
        while ($row = $result->fetch_assoc()) {
            $rankings[] = $row;
        }
        echo json_encode(array($tableName => $rankings));
    }
} catch (Exception $e) {
    echo "Error ", $e->getMessage();
}

Answer 2

改变这个：

while ($row = $result->fetch_assoc()) {
    $rankings[$i] = $row;
    $i++;
}
echo json_encode(array($tableName => $rankings));

进入这个：

while ($row = $result->fetch_array(MYSQL_ASSOC)) {
    $rankings[$i] = $row;
    $i++;
}
echo json_encode($rankings);

用排名列表填充json数组

2 个答案: