PHP - Table id the same

Question

如何将相同的列(在这种情况下按"level"排序)放在一起?我做了一个高分，我从我的数据库中按级别列出他们。如果它们是相同的级别，我希望它们具有相同的ID。

但是我不想显示其他的ID。只有第一个。下面是一个例子:

ID - Name - Level
1 - John - 5
2 - David - 4
3 - Josh - 3
   - Sam - 3
4 - George - 2

所以我想把它们放在一起，但如果它们具有相同的级别，则只有第一个显示ID。

我不希望它看起来像:

1 - John - 5
2 - David - 4
3 - Josh - 3
3 - Sam - 3
4 - George - 2

现在，它只是列出每个人，并给每个人一个唯一的ID。即使他们有相同的"水平"。我该如何解决这个问题?下面是我的代码:

<?php
            $sql = mysql_query("SELECT * FROM rookstayers ORDER BY level DESC LIMIT 0, 500");
            $id = 1;
            while($row = mysql_fetch_array($sql)){
                $name = $row['name'];
                $level = $row['level'];
                $world = $row['world'];
                $account = $row['accountstatus'];
                $status = $row['onlinestatus'];
                $onrow = '';
                $typeServ = '';
                $player_name = urlencode($name);
                if ($status == 1){
                    $status = 'Online';
                    $onrow = 'online';
                } else {
                    $status = 'Offline';
                    $onrow = 'offline';
                }
                if ($account == 'Premium Account'){
                    $account = 'Premium';
                } else {
                    $account = 'Free';
                }
                if ($world == 'Aurora' || $world == 'Aurera'){
                    $typeServ = 'active';
                } else {
                    $typeServ = '';
                }
            echo "<tr class=" . $typeServ . ">";
             echo "<td>" . $id . "</td>";
              echo "<td>" . $name . " <a href='http://www.tibia.com/community/?subtopic=characters&name=" . $player_name . "' target='_blank'><img src='../../img/websites/tibia.png' title='Tibia Profile'></a><a href='http://www.pskonejott.com/otc_display.php?character=" . $player_name . "' target='_blank'><img src='../../img/websites/pskonejott.png' title='Pskonejott'></a>" . "</td>";
              echo "<td>" . $level . "</td>";
              echo "<td>" . $world . "</td>";
              echo "<td>" . $account . "</td>";
              echo "<td class=" . $onrow . ">" . $status . "</td>";
            echo "</tr>";
            $id++;
}
          echo "</tbody>";
        echo "</table>";
?>

Answer 1

您可以为当前级别创建一个临时变量，并检查它是否会在输出中显示id。

    $id = 1;
     $last_player_lvl = '';
     while($row = mysql_fetch_array($sql)){
      ///....
     echo "<tr class=" . $typeServ . ">";
                         echo "<td>" . ( ($last_player_lvl == $row['level']) ? '' : $id ) . "</td>";
                          echo "<td>" . $name . " <a href='http://www.tibia.com/community/?subtopic=characters&name=" . $player_name . "' target='_blank'><img src='../../img/websites/tibia.png' title='Tibia Profile'></a><a href='http://www.pskonejott.com/otc_display.php?character=" . $player_name . "' target='_blank'><img src='../../img/websites/pskonejott.png' title='Pskonejott'></a>" . "</td>";
                          echo "<td>" . $level . "</td>";
                          echo "<td>" . $world . "</td>";
                          echo "<td>" . $account . "</td>";
                          echo "<td class=" . $onrow . ">" . $status . "</td>";
                        echo "</tr>";
if($last_player_lvl == $row['level']){
    $id = $id;
}else{
    $id++;
}
$last_player_lvl = $row['level'];  
//....