mysql - Selecting the best score per player

Question

I have a table called Scores which contains columns: id, player_id, value1, value2, value3 and date.

The table has next following content:

+------+-----------+--------+--------+--------+------------+
|  id  | player_id | value1 | value2 | value3 |    date    |
+------+-----------+--------+--------+--------+------------+
|   1  |     1     |   10   |    0   |   0    | 2012-08-02 |
+------+-----------+--------+--------+--------+------------+
|   2  |     2     |   15   |    1   |   0    | 2012-08-03 |
+------+-----------+--------+--------+--------+------------+
|   3  |     3     |    9   |    0   |   0    | 2012-08-04 |
+------+-----------+--------+--------+--------+------------+
|   4  |     1     |   11   |    0   |   0    | 2012-08-05 |
+------+-----------+--------+--------+--------+------------+
|   5  |     2     |   16   |    2   |   0    | 2012-08-06 |
+------+-----------+--------+--------+--------+------------+
|   6  |     2     |   15   |    0   |   0    | 2012-08-07 |
+------+-----------+--------+--------+--------+------------+

I am trying to get a query which returns the best highscore of each player ordered by the value in "value1, value2, value3". Value1 is the field with more importance, value2 medium importance and value3 minor importance, example:

value1 = 15                              value1 = 15
value2 = 1       is greater than ->      value2 = 0
value3 = 0                               value3 = 1

The expected result from the query which I need is:

+------+-----------+--------+--------+--------+------------+
|  id  | player_id | value1 | value2 | value3 |    date    |
+------+-----------+--------+--------+--------+------------+
|   5  |     2     |   16   |    2   |   0    | 2012-08-06 |
+------+-----------+--------+--------+--------+------------+
|   4  |     1     |   11   |    0   |   0    | 2012-08-05 |
+------+-----------+--------+--------+--------+------------+
|   3  |     3     |    9   |    0   |   0    | 2012-08-04 |
+------+-----------+--------+--------+--------+------------+

I'm trying with MAX, DISTINCT, GROUP BY and sub-queries but I don't get the correct result. Basically it is the next query but picking the first row of each "group":

SELECT id, player_id, value1, value2, value3
   FROM scores
   ORDER BY value1 DESC, value2 DESC, value3 DESC

------EDIT 1-------

eggyal's answer works fine but, maybe, the performance is not too good. I need to benchmark his solution against large database to check response times.

I have had an idea (and possible solution). The solution consists adding new boolean column which says if that score is the best score of that player or not. This way I need to check if the new score is better than the best old score of that player when I'm adding new score into DB, if it is I need to mark the flag as false in the old best score and as true in the new score. This gives me a way to retrieve the best score of each player directly (simple query like SELECT ... FROM .... ORDER BY).

------EDIT 2-------

weicap's answer is the fastest solution. I don't know why but his query is twice more faster than eggyal's query.

------EDIT 3------- I was wrong, weicap's query is more faster if the query was cached previously, if it wasn't the query takes ten or more seconds. In change, weicap's answer always takes 300-400ms against 80.000 rows.

Answer 1

ごとに、グループごとの最大値valueを取得できます。

SELECT * FROM Scores NATURAL JOIN (
  SELECT player_id, value1, value2, MAX(value3) value3 FROM Scores NATURAL JOIN (
  SELECT player_id, value1, MAX(value2) value2         FROM Scores NATURAL JOIN (
  SELECT player_id, MAX(value1) value1                 FROM Scores
    GROUP BY player_id) t
    GROUP BY player_id) t
    GROUP BY player_id) t
ORDER BY value1 DESC, value2 DESC, value3 DESC

sqlfiddleで参照してください。

Answer 2

あなたはこれを試すことができます

SELECT player_id,
  (SELECT value1
   FROM Scores b where a.player_id=b.player_id  ORDER BY value1 DESC, value2 DESC, value3 DESC limit 1) as value1,
  (SELECT value2
   FROM Scores b where a.player_id=b.player_id  ORDER BY value1 DESC, value2 DESC, value3 DESC limit 1) as value2,
  (SELECT value3
   FROM Scores b where a.player_id=b.player_id  ORDER BY value1 DESC, value2 DESC, value3 DESC limit 1) as value3

FROM Scores a GROUP BY player_id order by value1 DESC, value2 DESC, value3 DESC

または何かのような

SELECT * FROM Scores a 

where id =(SELECT id
   FROM Scores b where a.player_id=b.player_id  ORDER BY value1 DESC, value2 DESC, value3 DESC limit 1)

GROUP BY player_id order by value1 DESC, value2 DESC, value3 DESC

Answer 3

インデックスを追加して(player_id, value1, value2, value3)から、次のクエリを試してください。

SELECT
    s.*
FROM 
        Scores AS s
    JOIN
        ( SELECT DISTINCT
              player_id
          FROM 
              Scores
        ) AS p
      ON s.id =
         ( SELECT
               id
           FROM 
               Scores AS b
           WHERE
               b.player_id = p.player_id
           ORDER BY
               value1 DESC, value2 DESC, value3 DESC
           LIMIT 1
         ) 
ORDER BY
    s.value1 DESC, s.value2 DESC, s.value3 DESC ; 

Answer 4

あなたの実際のスコアは: 1000、1510、900 など... わかりましたか? 1 桁の位置が重要な 10 進数のようなものです。3 つの値を 1 つの値に変換してグループ化し、オンザフライで (クエリで) 見積もることも、(テーブルに書き込む前に) 事前に計算することもできます。

Answer 5

どの言語を使用していますか (SQL 以外)? SQL でソートできれば最も簡単ですが、これは不可能だと思います。PHP を使用している場合は、for ループに配置し、「bestScore」などのインデックスを配列に追加してから、次のように各スコアを個別に確認できます。

//Extract Data Here
for($outI=0;$outI<count($scores);$outI++){
    $scores[$outI]['bestScore'] = 0;
    for($innI=0;$innI<=3;$innI++){
        if ($scores[$outI]['value' . $innI+1] > $scores[$outI]['bestScore'])
            $scores[$outI]['bestScore'] = $scores[$outI]['value' . $innI+1];
    }
    echo 'The best score for ' . $scores[$outI]['player_id'] . ' was ' . $scores[$outI]['bestScore'] . '.<br />';
}

期待どおりに機能する場合は、各プレーヤーの最高のスコアが一覧表示されます。

Answer 6

痛みの最大値が 16 であると仮定して、これを試してください。

Select Scores.* from Scores, (SELECT player_id,max(17*17*value1+17*value2+value3) 
as max_score  FROM Scores group by player_id)t where 
(17*17*value1+17*value2+value3) = t.max_score and Scores.player_id=t.player_id

Answer 7

SELECT * FROM (
SELECT id, player_id, value1, value2, value3, `date`
   FROM scores
   ORDER BY value1 DESC, value2 DESC, value3 DESC
) x 
GROUP BY player_id
order by value1 DESC, value2 DESC, value3 DESC