如何从相乘结果中选择随机唯一值

Question

我有一个系统，用户可以通过不同类型的贡献获得 1 个或多个积分。这些存储在 2 个表中：

CREATE TABLE user_contribution_types (
  type_id INTEGER UNSIGNED NOT NULL AUTO_INCREMENT,
  title VARCHAR(255) NOT NULL,
  credits DECIMAL(5,2) UNSIGNED NOT NULL,
  valid TINYINT(1) UNSIGNED NOT NULL DEFAULT 1,

  PRIMARY KEY (type_id)
);

CREATE TABLE user_contributions (
  user_id INTEGER UNSIGNED NOT NULL,
  type_id INTEGER UNSIGNED NOT NULL,
  create_date DATETIME NOT NULL,
  valid TINYINT(1) UNSIGNED NOT NULL DEFAULT 1,

  FOREIGN KEY (user_id)
    REFERENCES users(user_id),
  FOREIGN KEY (type_id)
    REFERENCES user_contribution_types(type_id)
);

我可以选择自特定日期以来获得的总积分：

SELECT SUM(credits) AS total
FROM   user_contribution_types AS a
JOIN   user_contributions AS b ON a.type_id = b.type_id
WHERE  b.create_date >= '2017-05-01 00:00:00'
       AND a.valid = TRUE
       AND b.valid = TRUE

同样，我可以为 b.user_id 添加匹配项，以查找该特定用户的总积分。

我想做的是将获得的每个积分视为赠品的一个条目，并从总数中选择 3 个随机（唯一）user_ids。因此，如果一位用户获得了 26 个积分，他们将有 26 次获胜的机会。

如何使用 SQL 完成此操作，或者在应用程序级别执行此操作是否更有意义？我更喜欢尽可能接近真正随机的解决方案。

Answer 1

您可以通过计算累积分布并使用rand()来选择一个用户：

SELECT uc.*
FROM (SELECT uc.user_id, (@t := @t + total) as running_total
      FROM (SELECT uc.user_id, SUM(credits) as total
            FROM user_contribution_types ct JOIN
                 user_contributions c
                 ON ct.type_id = c.type_id
            WHERE c.create_date >= '2017-05-01' AND ct.valid = TRUE AND c.valid = TRUE
            GROUP BY uc.user_id
           ) uc CROSS JOIN
           (SELECT @t := 0) params
      ORDER BY rand()
     ) uc
WHERE rand()*@t BETWEEN (running_total - total) AND running_total;

如果rand() 正好在边界上，这将返回两个值的可能性很小。出于您的目的，这不是问题；你可以添加limit 1。

要将其扩展到多行，您只需将WHERE 子句修改为：

WHERE rand()*@t BETWEEN (running_total - total) AND running_total OR
      rand()*@t BETWEEN (running_total - total) AND running_total OR
      rand()*@t BETWEEN (running_total - total) AND running_total

问题是所有结果值可能都是相同的结果。

您可以随机选择三个以上的值。我倾向于选择更大的数字，比如 9：

WHERE 0.1*@t BETWEEN (running_total - total) AND running_total OR
      0.2*@t BETWEEN (running_total - total) AND running_total OR
      0.3*@t BETWEEN (running_total - total) AND running_total OR
      . . .
ORDER BY rand()  -- redundant, but why not?
LIMIT 3

或者更简单地说：

WHERE FLOOR( 10*(running_total - total)/@t)) <> FLOOR( 10*running_total/@t)
ORDER BY rand()
LIMIT 3

这更容易，因为您可以更改 10 并沿累积分布测试任意数量的等距点。

Answer 2

好吧，我无法让 Gordon 的代码在没有错误的情况下运行，所以我最终回到应用程序逻辑并遵循解决方案 found here。示例：

// pick a random winner since a given date
// optionally exclude certain users
public function getWinner($date, array $exclude = []) {
    if (!empty($exclude)) {
        $in = implode(',', array_fill(0, count($exclude), '?'));
        array_unshift($exclude, $date);

        $sql = "SELECT   b.user_id, SUM(credits) AS total
                FROM     user_contribution_types AS a
                JOIN     user_contributions AS b ON a.type_id = b.type_id
                WHERE    b.create_date >= ?
                         AND b.user_id NOT IN ($in)
                         AND a.valid = TRUE
                         AND b.valid = TRUE
                GROUP BY b.user_id";
        $sth = $this->db->prepare($sql);
        $sth->execute($exclude);
    } else {
        $sql = "SELECT   b.user_id, SUM(credits) AS total
                FROM     user_contribution_types AS a
                JOIN     user_contributions AS b ON a.type_id = b.type_id
                WHERE    b.create_date >= :date
                         AND a.valid = TRUE
                         AND b.valid = TRUE
                GROUP BY b.user_id";
        $sth = $this->db->prepare($sql);
        $sth->execute([':date' => $date]);
    }

    $result = [];
    while ($row = $sth->fetch(PDO::FETCH_ASSOC)) {
        $result[$row['user_id']] = floor($row['total']);
    }

    // cryptographically secure pseudo-random integer, otherwise fallback
    $total = array_sum($result);
    if (function_exists('random_int')) {
        $rand = $total > 0 ? random_int(0, $total - 1) : 0;
    } else {
        // fallback, NOT cryptographically secure
        $rand = $total > 0 ? mt_rand(0, $total - 1) : 0;
    }

    $running_total = 0;
    foreach ($result as $user_id => $credits) {
        $running_total += $credits;
        if ($running_total > $rand) {
            // we have a winner
            return $user_id;
        }
    }

    return false;
}

所以我基本上可以多次执行此代码，以选择多个获胜者：

$ts = '2017-01-01 00:00:00';
$first_place = getWinner($ts);
$second_place = getWinner($ts, [$first_place]);
$third_place = getWinner($ts, [$first_place, $second_place]);

除非发布替代解决方案，否则我将接受此作为答案。