使用MySQL数据创建带有下拉列表的表单 - Creating a form with drop downs using MySQL data

Creating a form with drop downs using MySQL data

当前在输出使用mysql数据的表单时遇到问题。这可能是因为我放置了代码，因为这对我来说都是新的。当前输出（更新）如下所示：http://gyazo.com/18ee007c6e3618245706398d5a1e0ed3

输出应该显示为：下拉菜单旁边的事件（例如跳远A）。此外，它目前只显示1（或2）个循环——它应该显示大约25个事件。如果有人能理解为什么它没有显示所有事件，那么请建议纠正这一点。（代码已更新）

echo "<form method ='"POST'">";
  if ($result = $con->query("SELECT Event.Name FROM event")) 
  {                                     
        while ($row = $result->fetch_assoc()) 
        {               
           echo $row['Name'] . ' '; // . ' <br> ';          
           if ($result = $con->query("SELECT Student.Form, Teacher.Form, Student.Forename, Student.Surname, Student_ID " .
                                "FROM Student, Teacher " .
                                    "WHERE Student.Form = Teacher.Form AND Teacher.Form = 'C'")) 
           {                        
              if ($result->num_rows) 
              {             
                  echo "<select name ='Student_ID'>";                       
                  while ($row = $result->fetch_assoc()) 
                  {                                     
                     echo "<option value ='" . $row['Student_ID'] . "'>" . $row['Forename'] . ' ' . $row['Surname'] . "</option>";
                  }
                  echo "</select>";                 
              }                                                                 
           }                
        }
   }        
echo "</form>";

尝试重命名第二个$row

while ($row = $result->fetch_assoc()) {

至

while ($row_1 = $result->fetch_assoc()) {

并使用$row_1

<select>标记置于while循环之外。

echo "<select>";
while (){
//do stuff
}
echo "</select>";