Question

我有一个系统，用户可以为不同类型的贡献获得1个或多个学分。它们存储在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，其中包含以下内容：

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添加一个匹配项，以查找该特定用户的总积分。

我想要做的是将获得的每个学分视为赠品的入口，并从总数中选择3个随机（唯一）user_id。因此，如果一个用户获得26个学分，他们将有26个获胜机会。

如何使用SQL完成此操作，还是在应用程序级别更有意义？我宁愿选择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，那么这将返回两个值的可能性微乎其微。为了您的目的，这不是问题;您只需添加limit 1。

要将此扩展为多行，您只需将WHERE子句修改为：

即可

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是相同的结果。

您可以随机选择三个以上的值。我的倾向是选择一个更大的数字，例如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

或更简单：

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间距点。

Answer 2

好吧，我无法让Gordon的代码无错误地运行，所以我最终还原到应用程序逻辑并遵循解决方案found here。例如：

// 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;
}

因此我基本上可以执行此代码，因为我想选择多个获胜者：

$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。

如何从乘法结果中选择随机唯一值

2 个答案: