在mysql php中循环和连接输出

Question

我有一个运行数据库的代码，并将所有内容输出到php页面。

$avg = mysql_query("SELECT subject, gradeone, gradetwo, gradethree, ((gradeone + gradetwo + gradethree) / 3) as average FROM grades");


$q = mysql_query("SELECT * FROM newstudent AS n JOIN grades AS g ON n.id = g.id ORDER BY n.id") or die (mysql_error());

$last_student = null;

while ($row = mysql_fetch_assoc($q))
    {
        if ($row['id'] !== $last_student)
        {
            $last_student = $row['id'];
            echo "Student ID: ".$row['id']."<br/>";
            echo "First Name: ".$row['firstname']."<br/>";
            echo "Last Name: ".$row['lastname']."<br/>";
            echo "Email: ".$row['email']."<br/>";
            echo "<br/>";
        }
            print "<table id=reporttable>"; 

            print "<tr id=toprow> <td>subject</td> <td>gradeone</td> <td>gradetwo</td> <td>gradethree</td> <td>average</td></tr>"; 

            print   "<tr>";          

                print   " <td>";    
                    print   $row["subject"];
                print   "</td>"  ;

                print   " <td>";    
                    print   $row["gradeone"];
                print   "</td>"  ;

                print   " <td>";    
                    print   $row["gradetwo"];
                print   "</td>"  ;

                print   " <td>";    
                    print   $row["gradethree"];
                print   "</td>"  ;

                while ($r = mysql_fetch_array($avg))
                {
                    print   " <td>";    
                        print   $r['average'];
                    print   "</td>"  ;
                }

            print   " </tr>";   
            print "</table>";             

}?>

期望的结果应如下所示

以下代码的结果很好，但只有一个小问题。

第二个while循环假设是计算平均值并输出新行的每个记录。相反，它做到了这一点：

任何人都知道如何确保每个平均成绩都与学生的每一行一致吗？

Answer 1

$q = mysql_query("
    SELECT
        n.id,
        n.firstname, 
        n.lastname,
        n.email,
        g.gradeone, 
        g.gradetwo, 
        g.gradethree, 
        ((g.gradeone + g.gradetwo + g.gradethree) / 3) AS average
    FROM
        newstudent n JOIN grades g USING (id)
    ORDER BY
        n.id
") or die (mysql_error());

尝试使用此查询，然后在一个循环中输出结果 - 删除

while ($r = mysql_fetch_array($avg))

带括号。

留下这样的东西：

....
print "<td>$row['gradetwo']</td>";
print "<td>$row['gradethree']</td>";

print "<td>$row['average']</td>";
....