从结果表生成体育联赛表答案

【问题标题】：generating sports league table from results table从结果表生成体育联赛表
【发布时间】：2015-04-28 22:12:34
【问题描述】：

我有一张表格，其中包含一个联赛中所有球队的赛程和结果。我正在尝试从结果中生成一个排名表。我以为我已经成功了，但是在手动计算排名时它与 MySql 输出的表不匹配。

$query_away = " Select teams.team_name,
    SUM(if(fixtures.away_team_score > fixtures.home_team_score,3,0)) AS W,
    SUM(IF(fixtures.away_team_score = fixtures.home_team_score,1,0)) AS D,
    SUM(IF(fixtures.away_team_score < fixtures.home_team_score,0,0)) AS L
FROM teams
 INNER JOIN fixtures ON teams.team_name = fixtures.home_team 
 GROUP BY fixtures.home_team
 ORDER BY W DESC";

似乎将不劳而获的 3 分分配给没有获胜的球队。有没有更简单的方法来实现这一点或修复我拥有的代码？总而言之，我试图计算客队得分主队的次数，并为此分配 3 分。与对手打平得 1 分，失败得 0 分。

小提琴http://sqlfiddle.com/#!9/85813/1

编辑

此查询重复两次，一次用于主场排名，一次用于客场排名。加入away_team 上的客场查询修复了不劳而获的 3 分问题，但如果我能从一个查询中获得排名，这将有所帮助。代码如下。

$fullTable = [];
$sortedTable = [];

$query_away = " Select teams.team_name,
    SUM(if(fixtures.away_team_score > fixtures.home_team_score,3,0)) AS W,
    SUM(IF(fixtures.away_team_score = fixtures.home_team_score,1,0)) AS D,
    SUM(IF(fixtures.away_team_score < fixtures.home_team_score,0,0)) AS L
 FROM teams
 INNER JOIN fixtures ON teams.team_name = fixtures.away_team 
 GROUP BY fixtures.away_team
 ORDER BY W DESC";



$query_home = " Select teams.team_name, 
    SUM(if(fixtures.home_team_score > fixtures.away_team_score,3,0)) AS W,
    SUM(IF(fixtures.home_team_score = fixtures.away_team_score,1,0)) AS D,
    SUM(IF(fixtures.home_team_score < fixtures.away_team_score,0,0)) AS L
 from teams
 inner join fixtures on teams.team_name = fixtures.home_team 
 GROUP BY fixtures.home_team
 order by W desc";


$home_result = mysqli_query($dbc, $query_home);
$away_result = mysqli_query($dbc, $query_away);
echo'<table><tr><th>Home Table</th><th>W</th><th>D</th><th>L</th><th>Pts</th></tr>';
if (!$home_result) {
    echo 'no result';
} else {
    //print_r(mysqli_fetch_array($result));

    while ($row = mysqli_fetch_array($home_result)) {
        $pts = $row['W'] + $row['D'];
        echo "<tr><td>" . $row['team_name'] . "</td><td>" . $row['W'] / 3 . "</td><td>" . $row['D'] . "</td><td>" . $row['L'] . "</td><td>" . $pts . "</td><tr>";
        $homeTeam = $row['team_name'];
    $fullTable["$homeTeam"] = $pts;
}
    echo'</table>';
}

【问题讨论】：

带有示例数据的sql fiddle 会有所帮助。
您正在加入 teams.team_name = fixtures.home_team 上的团队表。这意味着当您执行if away_team_score > home_team_score 时，您将为失败的团队增加 3。
@Dagon fiddle 添加见 Q.
@Don'tPanic 你能解释一下吗
我认为如果你加入 team_name = away_team，你应该会得到你期望的结果。

标签： php mysql select join count

【解决方案1】：

这个查询应该同时获得主客场胜利三分。平局只需要一笔金额，因为无论球队是主场还是客场都无关紧要。损失不需要总和，因为无论如何你只是把一堆零加起来。

SELECT teams.team_name,
    SUM(if(teams.team_name = fixtures.away_team 
        AND fixtures.away_team_score > fixtures.home_team_score,3,0)) 
    + SUM(if(teams.team_name = fixtures.home_team 
        AND fixtures.home_team_score > fixtures.away_team_score,3,0)) AS W,
    SUM(IF(fixtures.away_team_score = fixtures.home_team_score,1,0)) AS D,
    0 AS L
FROM teams
INNER JOIN fixtures ON teams.team_name = fixtures.home_team 
    OR teams.team_name = fixtures.away_team
GROUP BY teams.team_name
ORDER BY W DESC

【讨论】：