2

我正在尝试计算一个团队在有序 MySQL 结果集中的排名，而我遇到的问题是检测第一个团队的平局以显示平局值。

例如，假设结果集如下：

team_id    pts
---------------
1          89
2          87
3          76
4          76
5          52

我用以下 PHP 计算团队的排名：

$i = 0;
while($row = mysql_fetch_assoc($r)) { //iterate thru ordered (desc) SQL results
    ++$i;
    ($row['pts'] == $prev_val)
        ? $rnk = 'T-' . $rnk    //same as previous score, indicate tie
        : $rnk = $i;            //not same as previous score
    $rnk = str_replace('T-T-','T-',$rnk); //eliminate duplicative tie indicator
    if ($row['team_id'] == $team_id) { //current team in resultset matches team in question, set team's rank
        $arr_ranks['tp']['cat'] = 'Total Points';
        $arr_ranks['tp']['actual'] = number_format($row['pts'],1);
        $arr_ranks['tp']['league_rank'] = $rnk;
        $arr_ranks['tp']['div_rank'] = $div_rnk;
    }
    else if ($i == 1) { //current team is category leader (rank=1) and is not team in question, set current team as leader
        $arr_ranks['tp']['leader'] = "<a href='index.php?view=franchise&team_id=" . $row['team_id'] . "'>" . get_team_name($row['team_id']) . '</a> (' . number_format($row['pts'],1) . ')';
    }
    $prev_val = $row['pts']; //set current score as previous score for next iteration of loop
}

上面的“平局”逻辑将把＃4队与＃3队平局，但反之则不然。

换句话说，对于#3 队，$rnk = 3，而对于#4 队，$rnk = T-3。（两者都应该是“T-3”。）

所以问题就变成了：我如何在遍历结果时“向前看”以找出当前分数是否与列表后面的分数相同/重复，这样我就可以将它与随后的欺骗一起视为并列？

谢谢。

编辑：如果我首先将结果存储在表中，我可以实现 Ignacio 的代码，"wins"如下所示：

select
    s1.team_id,
    t.division_id,
    sum(s1.score>s2.score) tot_wins,
            (   select count(*)
                from wins
                where team_id <> s1.team_id
                and wins > (select wins
                            from wins
                            where team_id = s1.team_id)
            )+1 as rnk
from
    scoreboard s1
    left join teams t
        on s1.team_id = t.team_id
    left join scoreboard s2
        on s1.year=s2.year and s1.week=s2.week and s1.playoffs=s2.playoffs and s1.game_id=s2.game_id and s1.location<>s2.location
group by s1.team_id
order by tot_wins desc;

这给出了以下结果：

team_id    division_id   tot_wins  rnk
--------------------------------------
10         1             44        1
2          1             42        2
3          2             42        2
8          1             39        4
5          2             37        5
. . .

但是，我突然想到我已经得到了这个结果集，这实际上并没有解决我的问题。

为避免混淆，我已经单独发布了“跟进”问题。

4

3 回答 3

2

我喜欢伊格纳西奥与他的答案的链接。但是，如果您仍然想使用 PHP，您可以按 SCORE 收集排名，并为每个分数分配团队。这可能不是最有效的方法，但它会起作用。

$ranks = array();
while ($row = mysql_fetch_assoc($result)) {
    $ranks[$row['pts']][] = $row['team_id'];
}

$ranks将是一个看起来像...的数组

$ranks[89] = array(1);
$ranks[87] = array(2);
$ranks[76] = array(3,4);
$ranks[52] = array(5);

使用foreachon $ranks，并仔细检查点的出现方式（升序或降序）。您可以使用 count() 查看是否有平局。

于 2012-06-01T20:40:39.387 回答
1 
$exists = array();
if ($row['team_id'] == $team_id && !in_array($row['pts'], $exists)) { //current team in resultset matches team in question, set team's rank
        $exists[] = $row['pts'];
        $arr_ranks['tp']['cat'] = 'Total Points';
        $arr_ranks['tp']['actual'] = number_format($row['pts'],1);
        $arr_ranks['tp']['league_rank'] = $rnk;
        $arr_ranks['tp']['div_rank'] = $div_rnk;
    }
于 2012-06-01T18:36:01.973 回答
1

这个问题已经在这里回答了。

查询：

SELECT a.team_id, a.wins, count(*) instances
FROM
    (SELECT
        s1.team_id,
        sum(s1.score>s2.score) wins
    FROM scoreboard s1
        LEFT JOIN scoreboard s2
            ON s1.year=s2.year
            AND s1.week=s2.week
            AND s1.playoffs=s2.playoffs
            AND s1.game_id=s2.game_id
            AND s1.location<>s2.location
    GROUP BY s1.team_id) AS a
    LEFT JOIN
        (SELECT
            sum(s1.score>s2.score) wins
        FROM scoreboard s1
            LEFT JOIN scoreboard s2
                ON s1.year=s2.year
                AND s1.week=s2.week
                AND s1.playoffs=s2.playoffs
                AND s1.game_id=s2.game_id
                AND s1.location<>s2.location
        GROUP BY s1.team_id) AS b
            ON a.wins = b.wins
GROUP BY a.team_id, b.wins
ORDER BY a.wins DESC;

这给出了输出......

=================================
|team_id   | wins    |instances |
=================================
|10        | 44      |1         |
|2         | 42      |3         | //tie
|9         | 42      |3         | //tie
|5         | 42      |3         | //tie
|3         | 41      |1         |
|11        | 40      |1         |
|...       |         |          |
=================================

然后，在 PHP 中，我将能够通过检查 when 来检测所有关系$row['instances'] > 1。

于 2012-06-13T18:15:16.723 回答