1 つのオプションは、次のようなランキング変数を使用することです。
UPDATE player
JOIN (SELECT p.playerID,
@curRank := @curRank + 1 AS rank
FROM player p
JOIN (SELECT @curRank := 0) r
ORDER BY p.points DESC
) ranks ON (ranks.playerID = player.playerID)
SET player.rank = ranks.rank;
このパーツでは、別のコマンドJOIN (SELECT @curRank := 0)
を必要とせずに変数を初期化できます。SET
このトピックに関する詳細情報:
テストケース:
CREATE TABLE player (
playerID int,
points int,
rank int
);
INSERT INTO player VALUES (1, 150, NULL);
INSERT INTO player VALUES (2, 100, NULL);
INSERT INTO player VALUES (3, 250, NULL);
INSERT INTO player VALUES (4, 200, NULL);
INSERT INTO player VALUES (5, 175, NULL);
UPDATE player
JOIN (SELECT p.playerID,
@curRank := @curRank + 1 AS rank
FROM player p
JOIN (SELECT @curRank := 0) r
ORDER BY p.points DESC
) ranks ON (ranks.playerID = player.playerID)
SET player.rank = ranks.rank;
結果:
SELECT * FROM player ORDER BY rank;
+----------+--------+------+
| playerID | points | rank |
+----------+--------+------+
| 3 | 250 | 1 |
| 4 | 200 | 2 |
| 5 | 175 | 3 |
| 1 | 150 | 4 |
| 2 | 100 | 5 |
+----------+--------+------+
5 rows in set (0.00 sec)
更新:同じランクを共有するには同点が必要であることに気付きました。これは少しトリッキーですが、さらに多くの変数を使用して解決できます。
UPDATE player
JOIN (SELECT p.playerID,
IF(@lastPoint <> p.points,
@curRank := @curRank + 1,
@curRank) AS rank,
@lastPoint := p.points
FROM player p
JOIN (SELECT @curRank := 0, @lastPoint := 0) r
ORDER BY p.points DESC
) ranks ON (ranks.playerID = player.playerID)
SET player.rank = ranks.rank;
テスト ケースとして、175 ポイントの別のプレーヤーを追加してみましょう。
INSERT INTO player VALUES (6, 175, NULL);
結果:
SELECT * FROM player ORDER BY rank;
+----------+--------+------+
| playerID | points | rank |
+----------+--------+------+
| 3 | 250 | 1 |
| 4 | 200 | 2 |
| 5 | 175 | 3 |
| 6 | 175 | 3 |
| 1 | 150 | 4 |
| 2 | 100 | 5 |
+----------+--------+------+
6 rows in set (0.00 sec)
また、同点の場合にランクをスキップする必要がある場合は、別のIF
条件を追加できます。
UPDATE player
JOIN (SELECT p.playerID,
IF(@lastPoint <> p.points,
@curRank := @curRank + 1,
@curRank) AS rank,
IF(@lastPoint = p.points,
@curRank := @curRank + 1,
@curRank),
@lastPoint := p.points
FROM player p
JOIN (SELECT @curRank := 0, @lastPoint := 0) r
ORDER BY p.points DESC
) ranks ON (ranks.playerID = player.playerID)
SET player.rank = ranks.rank;
結果:
SELECT * FROM player ORDER BY rank;
+----------+--------+------+
| playerID | points | rank |
+----------+--------+------+
| 3 | 250 | 1 |
| 4 | 200 | 2 |
| 5 | 175 | 3 |
| 6 | 175 | 3 |
| 1 | 150 | 5 |
| 2 | 100 | 6 |
+----------+--------+------+
6 rows in set (0.00 sec)
注: 私が提案しているクエリはさらに単純化できることを考慮してください。