如何選擇會話用戶的匹配值的非激情用戶在PHP

Question

我得到了兩個table一個是student和另一個是teacher所以基本上我想要一個選擇查詢這個表，如果某些教師教Comp Comp部門和此記錄被保存在teacher表中department列和另一種情況下某些學生也在同一部門學習和他的記錄也保存在student表department列所以我的問題，我只希望老師被選中誰的部門是平等的給登錄用戶

表student

studentID department 
     1  Comp Sci 
     2  Mech Engg 

表teacher

teacherID department name 
     1  Comp Sci  XYZ 
     2  Mech Engg ABC 

在我的代碼只從兩個表中選擇匹配的值，我不想這樣，我不知道該怎麼做

<?php 
     $sql=("SELECT * 
      FROM student 
       LEFT JOIN teacher ON student.studentID=teacher .teacher ID 
      WHERE student.department=teacher .department "); 
      $res=mysql_query($sql) or die (mysql_error()); 
      while($admin=mysql_fetch_assoc($res)) { 
      echo $admin['name']; 
      echo $admin['department']; 
      ?> 

Answer 1

SELECT * 
FROM 
    teacher 
INNER JOIN 
    student 
ON 
    student.department = teacher.department 
WHERE 
    student.studentID = $loginid // $login is the id of currently logged in student