php - 通过 subject_id 获取讲师姓名

Question

大家好，我的 mySQL 查询有问题。请查看我的表格和当前查询。

表“教师”

---------------------------------------------
id |firstname |     lastname    | subject_id|
---------------------------------------------
1    Juan         Dela Cruz            1
2    Antonio      Brigada              2
--------------------------------------------

表格“科目”

---------------
id |   name   |
---------------
1    English
2    Math
3    Science
--------------

我当前的查询

public function instructor(){
global $database;

$sql = "SELECT instructors.firstname as firstname, instructors.lastname as lastname ";
$sql .= "FROM subjects ";
$sql .= "JOIN instructors ";
$sql .= "WHERE subjects.id = instructors.subject_id";

$result = $database->query($sql);

while($instructor = $database->fetch_array($result)){
    if (isset($instructor['firstname']) && isset($instructor['lastname'])){
            return $instructor['lastname'] . " " . $instructor['firstname'];
    } else {
        return "N/A";
    }

    }
}

结果:

English    Juan Dela Cruz
Math       Juan Dela Cruz
Science    Juan Dela Cruz

编辑

当前的sql:

public function instructor(){
global $database;

$sql = "SELECT  a.name, ";
$sql .= "COALESCE(CONCAT(b.firstname, ' ',b.lastname), 'N/A') as ins_name ";
$sql .= "FROM subjects a ";
$sql .= "LEFT JOIN instructors b ";
$sql .= "ON a.id = b.subject_id";

$result = $database->query($sql);

   while($instructor = $database->fetch_array($result)){
      return $instructor['ins_name'];
   }
}

结果

English    Juan Dela Cruz
Math       Juan Dela Cruz
Science    Juan Dela Cruz

它应该是什么:

English    Juan Dela Cruz
Math       Antonio Brigada
Science    N/A

调用 instructor() 函数

<?php foreach($subjects as $subject): ;?>
  <tr class="subject_list">
    <td><?php echo $subject->name; ?></td> // This will show all subjects
    <td><?php echo $subject->instructor(); ?></td> // This will show instructor foreach subjects
  </tr>
<?php endforeach; ?>

希望大家帮我看看上面的结果

Answer 1

LEFT JOIN 是您所需要的

SELECT  a.name, 
        COALESCE(CONCAT(b.firstName, ' ',b.LastName), 'N/A')
FROM    subjects a
        LEFT JOIN instructors b
            ON a.ID = b.Subject_ID

• SQLFiddle Demo