php - MySQL数据库选择运动员数据需要位置

Question

我正在从数据库中选择数据并将其显示在页面上。
$position 是一个未保存在数据库中的有用值。 。显示每个运动员的位置很有用，如图所示，我只显示递增的 $i值(value)。

代码如下ORDERS BY run points 。问题是当按其他内容对列表进行排序时，例如 ORDER BY athlete ASC分配给每个运动员的位置不会随着运动员的移动而移动。

$selectproduct = mysqli_query($link, "SELECT `athlete`,`reference`,SUM(`Run Points`) AS 'Run Points',COUNT(`eventID`) AS 'num_runs',`Volunteer Points`,`location`,`Gender pos` FROM `Events` WHERE `Gender`='$Gender' AND `location`='$location' GROUP BY `reference` ORDER BY `Run Points` DESC");


while ($rowGetDetails = mysqli_fetch_array($selectproduct)){

$reference=$rowGetDetails['reference'];
$athlete=$rowGetDetails['athlete'];
$points=$rowGetDetails['Run Points'];
$num_runs=$rowGetDetails['num_runs'];
$position=$i;

echo '<tr>';
echo'<td>';
echo $athlete;
echo '</td>';
echo'<td>';
echo $num_runs;
echo '</td>';
echo'<td>';
echo $points;
echo '</td>';
echo '</tr>';

$i=$i+1;
}

按其他内容排序时，例如 athlete ASC在 $selectproduct它仍然给出 $postition列表中第一个名字以 A 开头的运动员如果不是得分最多的运动员，则为 1。

$selectproduct 还有其他选项例如按 athlete ASC 排序如下:

 $selectproduct = mysqli_query($link, "SELECT `athlete`,`reference`,SUM(`Run Points`) AS 'Run Points',COUNT(`eventID`) AS 'num_runs',`Volunteer Points`,`location`,`Gender pos` FROM `Events` WHERE `Gender`='$Gender' AND `location`='$location' GROUP BY `barcode` ORDER BY `athlete` ASC"

Answer 1

只需在循环外分配 $i 值即可；

$selectproduct = mysqli_query($link, "SELECT `athlete`,`reference`,SUM(`Run Points`) AS 'Run Points',COUNT(`eventID`) AS 'num_runs',`Volunteer Points`,`location`,`Gender pos` FROM `Events_08052017_withdate` WHERE `Gender`='$Gender' AND `location`='$location' GROUP BY `reference` ORDER BY `Run Points` DESC");
    $i = 1;

    while ($rowGetDetails = mysqli_fetch_array($selectproduct)){

    $reference=$rowGetDetails['reference'];
    $athlete=$rowGetDetails['athlete'];
    $points=$rowGetDetails['Run Points'];
    $num_runs=$rowGetDetails['num_runs'];
    $position=$i;

    echo '<tr>';
    echo'<td>';
    echo $athlete;
    echo '</a>';
    echo '</td>';
    echo'<td>';
    echo $num_runs;
    echo '</td>';
    echo'<td>';
    echo $points;
    echo '</td>';
    echo '</tr>';

    $i=$i+1;
    }

尝试查询这样的内容可能适合您

SET @rank=0;
SELECT @rank := @rank + 1 AS ranking, t.avg, t.name
  FROM(SELECT avg(students_signatures.score) as avg, students.name as name
FROM alumnos_materia
JOIN (SELECT @rownum := 0) r
left JOIN students ON students.id=students_signatures.id_student
GROUP BY students.name order by avg DESC) t