php - 在 mySql LEFT JOIN 语句中使用多个 ON 条件不起作用

Question

我正在尝试将我的两个表left join 在一起，并且在 ON 之后仅使用一个条件时它可以工作，但是当我尝试添加第二个条件时，我不要从 firstname 和 lastname 列中获取任何信息。我一直在这里查看其他答案，从我所看到的情况来看，我所做的一切都是正确的，所以我无法弄清楚为什么它不起作用..

这是我的尝试:

$stmt = 'SELECT courses.*, teachers.firstname, teachers.lastname ';
$stmt .= 'FROM courses LEFT JOIN teachers ON ';
$stmt .= 'courses.main_teacher = teachers.id AND courses.secondary_teacher = teachers.id';

$prep_stmt = $db->prepare($stmt);
$prep_stmt->execute();

while ($db_row = $prep_stmt->fetch(PDO::FETCH_ASSOC)) {
    echo '<tr>';
    echo '<td class="info-cell">' . $db_row['name'] . '</td>';
    echo '<td class="info-cell">' . $db_row['curriculum_url'] . '</td>';
    echo '<td class="info-cell">' . $db_row['rating_criteria'] . '</td>';
    echo '<td class="info-cell">' . $db_row['firstname'] . ' ' . $db_row['lastname'] . '</td>'; //Echoes main_teacher
    echo '<td class="info-cell">' . $db_row['firstname'] . ' ' . $db_row['lastname'] . '</td>'; //Echoes secondary_teacher
    echo '<td class="info-cell">' . $db_row['for_programme'] . '</td>';
    echo '<td class="info-cell">
              <button class="small-btns"><img src="../includes/img/edit.png"></button>
          </td>';
    echo '<td class="info-cell">
              <button class="small-btns"><img src="../includes/img/delete.png"></button>
          </td>';
    echo '</tr>';  
}

这是它的输出:

如您所见，当使用多个条件时，它不会从教师表返回任何结果，这是为什么？

编辑:

我的数据库中的类(class)表:

我的数据库中的教师表:

我想要的期望结果:

Answer 1

看来你需要加入 teachers 2次

select
c.*, 
t1.firstname as main_teacher_firstname, 
t1.lastname as main_teacher_lastname,
t2.firstname as secondary_teacher_firstname, 
t2.lastname as secondary_teacher_lastname
from courses c
left join teacher t1 on t1.id = c.main_teacher
left join teacher t2 on t2.id = c.secondary_teacher