我正在尝试计算有序
MySQL结果集中团队的排名,我遇到的问题是检测第一个团队出现绑定值的关系.
例如,假设结果集如下：
team_id pts --------------- 1 89 2 87 3 76 4 76 5 52
我使用以下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 = $row['pts']; //set current score as prevIoUs score for next iteration of loop
}
上面的“平局”逻辑将捕获团队#4与团队#3并列,但反之则不然.
换句话说,对于团队#3,$rnk = 3,而对于团队#4,$rnk = T-3. (两者都应该是“T-3”.)
所以问题就变成了：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,那么我可以将其作为一个平局与随后的欺骗一起对待？
谢谢.
编辑：如果我首先将结果存储在一个表中,我可以实现Ignacio的代码,例如下面的“wins”：
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,我发现我已经到了这个结果集,这实际上并没有解决我的问题.
为避免混淆,我发布了the “follow-up” problem separately.
我喜欢伊格纳西奥与他答案的联系.但是如果您仍然想使用PHP,您可以通过score收集排名并为每个分数分配团队.它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'];
}
$rank是一个看起来像……的数组.
$ranks[89] = array(1); $ranks[87] = array(2); $ranks[76] = array(3,4); $ranks[52] = array(5);
在$rank上使用foreach,并仔细检查积分将以哪种方式出现(升序或降序).您可以使用count()来查看是否存在平局.