内连接返回非预期的NULL (PHP/MySQL)

我目前正在运行一个函数，该函数在while循环中使用内部连接，该循环意外地不返回任何内容。

$data3 = '''' . implode(''', ''', $posted_email) . '''';
$exam_recordation = mysql_query("SELECT * FROM exam WHERE partner_id IN (SELECT partner_id FROM exam WHERE Student_email = $data3) AND Student_email <> $data3");
//$name_query = mysql_query("SELECT student.email, student.name, Student_email FROM student INNER JOIN exam ON student.email = exam.Student_email WHERE student.email = $data3")
echo "<table border=='1'>
<tr>
<th>Name</th>
<th>Email</th>
<th>Exam Date</th>
<th>Level</th>
<th>Mark</th>
    <th>Style</th>
</tr>";
while($row = mysql_fetch_array($exam_recordation))
  {
  echo "<tr>";
$name_query = mysql_query("SELECT DISTINCT student.name
              FROM student
              INNER JOIN exam ON student.email = exam.Student_email
              WHERE student.email <>{$data3}");
//die();
 /*while($row1 = mysql_fetch_array($name_query))
  {
    echo "<td>" . $row1['name'] . "</td>"; //This while loop was just me overcomplicating it, soon worked out I didnt need it
  }*/
  echo "<td>" . $name_query['name'] . "</td>";
  echo "<td>" . $row['Student_email'] . "</td>";
  echo "<td>" . $row['examDate'] . "</td>";
  echo "<td>" . $row['level'] . "</td>";
  echo "<td>" . $row['mark'] . "</td>";
      echo "<td>" . $row['style'] . "</td>";
  echo "</tr>";
  }
echo "</table>";

从我自己的尝试调试它，我已经意识到，它返回的不是我希望的值(学生表的名字)，而是一个NULL值，当我在一个单独的PHP函数中运行它自己，即不在下面显示的while循环内。

$data3 = '''' . implode(''', ''', $posted_email) . '''';
var_dump($posted_email);
var_export($data3);
$name_query = mysql_query("SELECT DISTINCT student.name
              FROM student
              INNER JOIN exam ON student.email = exam.Student_email
              WHERE student.email <> $data3");
var_export($name_query);
echo "SELECT DISTINCT student.name
              FROM student
              INNER JOIN exam ON student.email = exam.Student_email
              WHERE student.email <> $data3";

然而，当我从SQL中的最后一位代码运行echo'd INNER JOIN时，它工作良好并返回2行名称。

SELECT DISTINCT student.name FROM student INNER JOIN exam ON student.email = exam.Student_email WHERE student.email <> 'eating@gnomes.com'

我正在研究的理论是这样的:

$name_query = mysql_query("SELECT DISTINCT student.name
              FROM student
              INNER JOIN exam ON student.email = exam.Student_email
              WHERE student.email <>{$data3}");
while($row = mysql_fetch_array($exam_recordation) && mysql_fetch_array($name_query))
  {
  echo "<tr>";
  echo "<td>" . $row['name'] . "</td>";
  echo "<td>" . $row['Student_email'] . "</td>";
  echo "<td>" . $row['examDate'] . "</td>";
  echo "<td>" . $row['level'] . "</td>";
  echo "<td>" . $row['mark'] . "</td>";
      echo "<td>" . $row['style'] . "</td>";
  echo "</tr>";
  }
echo "</table>";

这返回什么都没有到我的表，但我要做的是使用两个查询在一个fetch，以便将它们的结果都放入$行，并能够提取相关数据。这听起来有点牵强，如果是这样的话，那就怪我缺乏经验吧。

我希望你能看到，我已经试图找出问题是什么，但没有成功。通常只是检查正在发送的查询是否正确，已经修复了这种问题，过去没有任何东西为我返回，但似乎不是为了这个问题。可能是我缺少的数组或while循环，可能不是。我真的没办法了。