Question

我已成功设法选择并返回具有相关成员的团队。在成员表中有以下列：score_1, score_2等......

我正在努力将这些整数值从表格中取出来与数据并列：

<?php
$sql = "SELECT t.team_name as team_name, GROUP_CONCAT(m.firstName, ' ', m.lastName) as team_members 
FROM members AS m
JOIN team_members AS tm
ON tm.member_id = m.member_id 
JOIN teams as t
on t.team_id = tm.team_id
WHERE t.dashboard_id = $dashboard_id AND t.team_id = $teamSelect
GROUP BY t.team_name"; 
if(!$result = $conn->query($sql)) {
    // die(printf("Errormessage: %s\n", $conn->error));
}
while($row = $result->fetch_assoc()){
    echo '<h2>Team Scores: <span class="teamNameTable">' . $row["team_name"] . '</span></h2><br>';
    $names = explode(',', $row['team_members']);
    echo '<div class="tableHeader">';
    echo '<div class="col">Name</div>';
    echo '<div class="col">SDO</div>';
    echo '<div class="col">DCTO</div>';
    echo '<div class="col">ED</div>';
    echo '<div class="col">CA</div>';
    echo '<div class="col">DHPT</div>';
    echo '<div class="col">IRT</div>';
    echo '<div class="col">GL</div>';
    echo '<div class="col">IL</div>';
    echo '</div>';
    foreach($names as $name) {
        echo '<div class="teamNameMember">' . $name  . '</div>';
    }   
    echo '<br>';  
} 
?>

更新

foreach($names as $name) {
echo '<div class="teamNameMember">' . $name  . '</div>';
}   
echo '<br>';  
}

Answer 1

您忘了将这些列添加到您的选择语句...

$sql = "SELECT t.team_name as team_name, GROUP_CONCAT(m.firstName, ' ', m.lastName) as team_members, m.score_1, m.score_2 
FROM members AS m
JOIN team_members AS tm
ON tm.member_id = m.member_id 
JOIN teams as t
on t.team_id = tm.team_id
WHERE t.dashboard_id = $dashboard_id AND t.team_id = $teamSelect
GROUP BY t.team_name";

选择其他表数据

1 个答案: