在迭代查询结果时识别重复项

时间：2012-06-12 16:51:47

标签： php mysql ranking

这是对question I asked previously。

的跟进

我可以计算等级（包括关系的逻辑）就好了;当我遇到重复排名的第一个实例时，问题是检测到未来的重复。

以下是获取结果集的SQL查询：

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
ORDER BY wins DESC;

以下是我将在PHP中循环的示例SQL结果集：

team_id   wins
--------------
10        52
2         48
5         46
11        46
3         42
9         39
...

这是我显示的PHP代码，需要将“T-”附加到所有绑定的等级：

$i = 0;
while($row = mysql_fetch_assoc($r)) { //iterate thru ordered (desc) SQL results
    ++$i;
    ($row['wins'] == $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 Wins';
        $arr_ranks['tp']['actual'] = number_format($row['wins'],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['wins']) . ')';
    }
    $prev_val = $row['wins']; //set current score as previous score for next iteration of loop
}

上面的“平局”逻辑将捕获团队＃4与团队＃3并列，但反之亦然。

换句话说，对于团队＃3，$rnk = 3，对于团队＃4，$rnk = T-3。 （两者应为“T-3”。）

所以问题变成了：如何在搜索结果的同时“向前看”以查明当前得分是否是列表中下方得分的平局/重复，因此我可以将其视为一个平局随后的欺骗？

@Airzooka给了我a potential solution，但我很想知道是否有更有效的方法（甚至可能在SQL级别）。

感谢。

5 个答案:

答案 0 :(得分：1)

在伪代码中：

loop through rows as row1
    loop through rows as row2
        if row1 ain't row2 and row1.points == row2.points, append T

更新

好的，这是怎么回事，因为你总是按胜利排序你的结果集：尝试在临时数组或变量中存储关于每一行的信息，比如$previousTeamWins， $previousTeamName等等。然后你可以比较当前和之前的，并根据它分配T.因此，您有效地将分配延迟到下一次迭代（或者直到最后一行的情况下循环退出）。通过行集的一次旅行应该完成工作。

答案 1 :(得分：0)

这个怎么样？

SELECT
 t1.*,
 EXISTS (
  SELECT *
  FROM `teams` as t2
  WHERE t2.pts = t1.pts AND t1.id != t2.id
 ) as `tied`
 FROM `teams` as t1
...

答案 2 :(得分：0)

试试这个我希望它为你工作

SELECT t1.`id`,t1.`pts`,t1.`team_id`,
IF(t1.`pts` = t2.`pts` AND t1.`id` != t2.`id` ,1,0) AS `tied`
FROM `teams` t1
LEFT JOIN `teams` t2 on t2.`pts` = t1.`pts` AND t2.`id` != t1.`id`
GROUP bY t1.`id`

答案 3 :(得分：0)

Oi，还在看你的代码。所以在一天结束时，你只想输出一两个团队，对吧？如果领导者是有问题的团队，那么就有一个，否则就是两个。如果是这种情况，请尝试这样的事情（警告它没有经过测试）：

$i = 0;
while($row = mysql_fetch_assoc($r)) { //iterate thru ordered (desc) SQL results 
    ++$i; 
    ($row['wins'] == $prev_wins) 
        ? $rnk = $prev_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 ($prev_team_id == $team_id) {
        $arr_ranks['tp']['cat'] = 'Total Wins'; 
        $arr_ranks['tp']['actual'] = number_format($prev_wins,1); 
        $arr_ranks['tp']['league_rank'] = $prev_rnk; 
        $arr_ranks['tp']['div_rank'] = $div_rnk;

        if ($row['wins'] == $prev_wins)
        {
            $arr_ranks['tp']['tie'] = true;
        }
        else
        {
            $arr_ranks['tp']['tie'] = false;
        }

    break;
    }
    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['wins']) . ')'; 
    }

    $prev_wins = $row['wins'];
    $prev_team_id = $row['team_id'];
    $prev_rnk = $rnk;
}

if ($prev_team_id == $team_id) {
    $arr_ranks['tp']['cat'] = 'Total Wins'; 
    $arr_ranks['tp']['actual'] = number_format($prev_wins,1); 
    $arr_ranks['tp']['league_rank'] = $prev_rnk; 
    $arr_ranks['tp']['div_rank'] = $div_rnk;

    if ($row['wins'] == $prev_wins)
    {
        $arr_ranks['tp']['tie'] = true;
    }
    else
    {
        $arr_ranks['tp']['tie'] = false;
    }

}

答案 4 :(得分：0)

我想我通过SQL找到了解决方案！可能不是很优雅（我可以稍后清理），但这里有......

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，我将通过检查$row['instances'] > 1的时间来检测所有联系。

感谢大家通过这个非常麻烦的问题与我联系！

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