PHP/SQL - 如果在foreach()中ID相同,则将值一起添加

Question

我试图显示玩家得分的统计数据,但是有时同一个玩家被添加到具有相同玩家ID的数据库中,当玩家被添加两次时我如何将这些值一起添加以显示为总数,而不是回声玩家两次.

db结构示例:

playerID | Goals | Season | leagueID
   1         5       1        1
   2         1       1        1
   1         2       1        2
   5         3       1        1
   1         3       2        2
   2         2       2        1

PHP:

$query = $db->query('SELECT * FROM playerstats ORDER BY goals DESC LIMIT 30');
$num = $query->num_rows;
if($num > 0) {

foreach($query as $row) {
$playerID = $row['playerID'];
$goals = $row['goals'];

echo '
<tr>
<td>'.$playerID.'</td>
<td>'.$goals.'</td>
</tr>

';

}

}

这将显示玩家ID 1,3分开的时间.如何让所有目标加在一起只显示一次playerID 1(10)

我已经尝试将查询更改为:SELECT DISTINCT*FROM playerstats ORDER BY目标DESC LIMIT 30但这没有任何区别.

Answer 1

Group BY将帮助您:

<?php
$query = $db->query('SELECT SUM(p.goals) as total_goals, 
p.playerID, p.leagueID, p.Season 
FROM playerstats as p 
GROUP BY p.playerID, p.leagueID, p.Season
ORDER BY total_goals DESC LIMIT 30');

$num = $query->num_rows;
if($num > 0) {
foreach($query as $row) {        
   echo '
    <tr>
    <td>'.$row['playerID'].'</td>
    <td>'.$row['total_goals'].'</td>
    </tr>

    ';

  }

}

请注意,我的查询也会根据赛季和联赛进行分组,如果你想要通过赛季和联赛的总进球,你的group by意志是: GROUP BY p.playerID

Answer 2

如果你只想要playerID和的总和goals不是无所谓的Season或leagueID更改查询到:

SELECT DISTINCT playerID AS player, (SELECT SUM(goals) FROM playerstats WHERE playerID = player) AS totalGoals FROM playerstats