Php "全胜";； $arr_排名['tp']['actual']=number_格式（$prev_wins，1）； $arr_ranks['tp']['league_rank']=$prev_rnk； $arr_ranks['tp']['div_rank']=$div_rnk； 如果（$row['wins']=$prev_wins） { $arr_秩['tp']['tie']=true； } 其他的 { $arr_秩['tp']['tie']=false； } 打破 } 否则，如果（$i==1）{//当前团队是类别领导（秩=1），并且不是有问题的团队，则将当前团队设置为领导 $arr_排名['tp']['leader']=“（'.number_格式（$row['wins']）”）”； } $prev_wins=$row['wins']； $prev_team_id=$row['team_id'； $prev_rnk=$rnk； } 如果（$prev_team_id==$team_id）{ $arr_排名['tp']['cat']='Total Wins'； $arr_排名['tp']['actual']=number_格式（$prev_wins，1）； $arr_ranks['tp']['league_rank']=$prev_rnk； $arr_ranks['tp']['div_rank']=$div_rnk； 如果（$row['wins']=$prev_wins） { $arr_秩['tp']['tie']=true； } 其他的 { $arr_秩['tp']['tie']=false； } }

我想我通过SQL找到了一个解决方案！可能不太优雅（我可以稍后再清理），但接下来

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;

然后，在PHP中，我将能够通过检查when

$row['instances']>1

来检测所有关系

---

感谢大家与我一起度过了这个不自然的麻烦问题！

我想我通过SQL找到了一个解决方案！可能不是很优雅（我可以稍后再清理），但接下来

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;

然后，在PHP中，我将能够通过检查when

$row['instances']>1

来检测所有关系

---

---

感谢大家帮助我解决了这个异常麻烦的问题！

为什么不在循环中一次性读取整个点列表，然后使用数组搜索检查下一个循环中的重复项？为什么不在循环中一次性读取整个点列表，然后使用数组搜索检查下一个循环中的重复项？Thanks，但这只有在结果存储在表中时才有效，而我的不是。我在上面的编辑中添加了我的查询，以便您可以看到我实际上必须计算值，然后检查这些值是否重复。您的结果不在表中？也许您应该删除巨大的标题，该标题为：这是获取数据的SQL查询e结果集：@BrianWarshaw我正在检查重复的值（“wins”）不在表中。正如您所看到的，我必须通过比较分数来计算它。谢谢，但这只有在结果存储在表中时才有效，而我的不是。我在上面的编辑中添加了我的查询，以便您可以看到，我实际上必须计算值，然后检查这些值是否重复。您的结果不在表中吗？也许您应该删除一个巨大的标题，它是：下面是获取结果集的SQL查询：@BrianWarshaw我正在检查重复的值（“wins”）不在表中。正如你所看到的，我必须通过比较分数来计算它。谢谢，但我相信我在上面对@Matmarbon的评论也适用于这里……谢谢，但我相信我在上面对@Matmarbon的评论也适用于这里……请看我在上面的评论。我必须通过查询来计算值；它们不是预先填充的在一个随时可用的表中。这很好——这个伪代码告诉您当您有行信息时该怎么做。我假设您已经了解PHP中的循环，所以它应该很容易实现。我认为[Airzooka提供的解决方案]（stackoverflow.com/questions/10855462/在php中迭代mysql结果时检测未来的重复值）将比两次循环结果更有效。我很好奇是否有比这更有效的方法，无论是直接通过SQL还是非直接通过SQL。他的效率稍高一些，是的，但只有在有关系的情况下——如果没有关系，你仍然有效地循环结果两次。那么你会说没有关系吗那么，在SQL级别提前检测关联的有效方法是什么呢？请参见我上面的评论。我必须通过查询来计算值；它们没有预先填充到一个现成的表中。这很好——这个伪代码告诉您当您有行信息时该怎么做。我假设您已经理解PHP中的循环，所以应该是这样的实现起来很简单。我认为[Airzooka提供的解决方案]（stackoverflow.com/questions/10855462/detect future duplicate value，同时在php中迭代mysql结果）将比两次循环结果更有效。我很好奇是否有比这更有效的方法，无论是直接通过SQL还是非直接通过SQL。他的效率稍高一些，是的，但只有在有关系的情况下——如果没有关系，你仍然有效地循环结果两次。那么你会说没有关系吗n在SQL级别提前检测联系的有效方法，那么？实际上，输出仅针对单个团队每页。输出是类别、实际值、排名（如果所讨论的团队与另一个团队联系在一起，则希望包括“T-”），和联赛领头羊。实际上，每页的输出仅针对单个球队。输出是类别、实际值、排名（如果所讨论的球队与其他球队并列，希望包括“T-”）和联赛领头羊。

$i = 0;
while($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
}

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`

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

$i = 0;
while($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;
    }

}

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;

=================================
|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         |
|...       |         |          |
=================================