PHP将一个查询中的值与另一个查询进行比较

我本质上是在尝试执行两个查询，一个是获取给定程序（或专业）的要求，另一个是获得学生已经完成的课程。

我想互相检查返回的字符串，如果用户参加了必修课，请在返回的要求旁边放一个"复选框"。如果他们没有参加必修课，请在要求旁边放一个空盒子。

我几乎已经达到了复选框概念的作用，然而，当我使用两个while循环时，它会重复返回的需求（课程）。

当我使用一个while循环时，它会获取第二个查询的数据并返回所有要求，但在找到第一个匹配项后就停止检查（例如，不继续查看课程2、3或4是否已经完成，等等）。

我还尝试了strcmp函数来检查$studentobject与$programsubject（programnumber和studentnumber同上）。

如果有人能提供如何实现这一点的煽动，或提供一种替代方法，我将不胜感激。如果需要，我可以提供更多细节。

         <table class="table">
    <tr>
            <th>&nbsp;</th>
        <th>Class</th>
    <th>Term</th>
    <th>Credits</th>
    </tr>

<?php 
$q = $db->query("SELECT * FROM `program_courses` a, `programs` b, `user_details` c WHERE a.pid = b.pid AND c.uid = '".$_GET['id']."' AND c.major = b.major");

while($program = $q->fetch()) {
$w = $db->query("SELECT * FROM `user_courses` WHERE uid = '".$_GET['id']."'");
$student = $w->fetch(); { // have also tried using a while loop here

$programsubject=$program['subject'];
$programnumber=$program['number'];

$studentsubject=$student['subject'];
$studentnumber=$student['number'];

?>

<?php 

if ($studentsubject==$programsubject && $studentnumber==$programnumber) {
        $checkbox = 'src="http://www.clipartbest.com/cliparts/ncX/jL6/ncXjL6rcB.png" width="25px" height="25px"';

    } elseif ($studentsubject!=$programsubject || $studentnumber!=$programnumber) {

        $checkbox = 'src="http://www.clker.com/cliparts/e/q/p/N/s/G/checkbox-unchecked.svg" width="25px" height="25px"';
    }
    ?>

    <?php
  //check off the requirement if the student has completed the course

        echo '
    <tr style="background-color:#E9FFD2">
    <td> <img '.$checkbox.'> </td>
            <td>'.$programsubject.' '.$programnumber.'</td>
    <td>&nbsp;</td>
    <td>3</td>

    </tr>';

?>

<?php
//End our conditionals
}
 }
 ?>
</table> 

编辑：提供了包括这些相应表格的表格结构在内的其他信息。

我希望避免更改两个相应的查询，但下面是每个查询返回的内容。我想将数据分开的原因是为了在未来引入更复杂的条件（例如，分数最低的课程可以计算在内，或者只有学分精确的课程才能计算在内，等等）。

对于用户1001，以下内容将转储。。。

SELECT * 
FROM `program_courses` a, `programs` b, `user_details` c 
WHERE a.pid = b.pid AND c.uid = '1001' AND c.major = b.major
id, pid, subject, number, credits, pid, major, title, degree, college, catalog, credits_req, id, uid, major, college, degree, catalog_year, credits, gpa, academic_standing, advisor, holds, id
'3','1','IDT','600','3','1','4574','Instructional Design & Technology','PHD','45','201408','72','1','1001','4574','45','Doctor of Philosophy','201408','52','3.44','Good Standing','Terence A','01'
'4','1','IDT','610','3','1','4574','Instructional Design & Technology','PHD','45','201408','72','1','1001','4574','45','Doctor of Philosophy','201408','52','3.44','Good Standing','Terence A','01'
'5','1','IDT','693J','3','1','4574','Instructional Design & Technology','PHD','45','201408','72','1','1001','4574','45','Doctor of Philosophy','201408','52','3.44','Good Standing','Terence A','01'
'6','1','IDT','750','3','1','4574','Instructional Design & Technology','PHD','45','201408','72','1','1001','4574','45','Doctor of Philosophy','201408','52','3.44','Good Standing','Terence A','01'
'7','1','IDT','790','3','1','4574','Instructional Design & Technology','PHD','45','201408','72','1','1001','4574','45','Doctor of Philosophy','201408','52','3.44','Good Standing','Terence A','01'
'8','1','IDT','691','3','1','4574','Instructional Design & Technology','PHD','45','201408','72','1','1001','4574','45','Doctor of Philosophy','201408','52','3.44','Good Standing','Terence A','01'
'21','1','IDT','931','3','1','4574','Instructional Design & Technology','PHD','45','201408','72','1','1001','4574','45','Doctor of Philosophy','201408','52','3.44','Good Standing','Terence A','01'
'22','1','IDT','660','3','1','4574','Instructional Design & Technology','PHD','45','201408','72','1','1001','4574','45','Doctor of Philosophy','201408','52','3.44','Good Standing','Terence A','01'

对于具有..的同一用户。。

SELECT * FROM `user_courses` WHERE id = '1001'
id, cid, uid, subject, number, credits, enroll_status, id, id
'1', '1', '1001', 'IDT', '610', '3', 'complete'
'3', '86903', '1001', 'IDT', '750', '3', 'complete'

因此，根据我的基本逻辑——只有当user_courses和program_courses表之间的subjectname和number匹配时，IDT 610和IDT 750才需要有一个完整的复选框，而所有其他程序需求都需要一个空的复选框。这有道理吗？到目前为止，我真的很感谢所有的反馈，因为在最佳实践方面，我不是一个大师。如果提供具有上述结构

if(in_array($program['subject'].$program['number'], $completed_courses)){
    // Yes the student has completed this course
}else{
    // no, the student has not completed this course
}