Php while-fetch数组|如何获取每个输出的代码？

我正在制作一个系统，我需要看到每个学生的信息。当我点击按钮时，我需要知道学生的代码并打开一个包含学生信息的页面（参见student.php）。有人能帮我吗

<?php

$sql = $conn->query("SELECT * from tb_coordinator inner join tb_teacher
on tb_coordinator.cd_coord = tb_teacher.cd_coord
inner join teacher_class on
tb_teacher.cd_teacher = teacher_class.cd_teacher
inner join tb_teacher on
teacher_class.cd_class = tb_class.cd_class 
inner join tb_student on
tb_class.cd_class = tb_student.cd_class
where tb_teacher.cd_cpf = '$cpf' and nm_class = '3a1'  order by cd_number asc");

while($row = mysqli_fetch_array($sql)){
    $nm_student = $row['nm_student'];
    $cd_number = $row['cd_number'];
?>


<ul class="list-group list-group-flush" style="width: 50%; margin-top: 2%;">
    <div id="load_student">
      <li class="list-group-item"><?php echo $cd_number . " - " . $nm_student;?>
       <a href="students.php" ><button type="button" class="btn btn-outline-info" style="margin-left: 90%;">Get Information</button></a></li>
    </div>
</ul>

</div>
<?php }?>

---

您需要对准备好的HTML进行一些修改。将您所需的参数添加到锚定
href
，这可能有助于您从
中的数据库获取学生详细信息，请参见student.php
页面

<ul class="list-group list-group-flush" style="width: 50%; margin-top: 2%;">
    <div id="load_student">
        <li class="list-group-item"><?php echo $cd_number . " - " . $nm_student;?>
            <a href="seeStudent.php?cd_number=<?=$cd_number?>&nm_student=<?=$nm_student?>" >
                <button type="button" class="btn btn-outline-info" style="margin-left: 90%;">Get Information</button>
            </a>
        </li>
    </div>
</ul>

<?php
    $cd_number = isset($_GET['cd_number']) ? $_GET['cd_number'] : '';
    $nm_student= isset($_GET['nm_student']) ? $_GET['nm_student'] : '';

    // .... code stuff, possibly query and display
?>