谁能告诉我如何获得 project_title、project_id、level_of_want 和 selection_id 的值。当我提交这个我得到没有选择数据库错误。
$query_Name = "SELECT u.Student_Surname, u.Student_Forename, p2.Project_Title,
p2.Project_id, s.level_of_want, s.selection_id
FROM users u
INNER JOIN projects p2 ON u.id = p2.Project_Lecturer
INNER JOIN selection s ON p2.Project_id = s.id_project
INNER JOIN users u2 ON s.student_id = u2.id
WHERE u2.Username = ".$_SESSION['MM_Username']." ORDER BY selection_id ASC" ;
$Name = mysql_query($query_Name, $projectsite) or die(mysql_error());
$row_Name = mysql_fetch_assoc($Name);
$totalRows_Name = mysql_num_rows($Name);