Php 表中的下拉列表

为什么我的下拉列表在表格中被分隔为不同的列。示例我希望在下拉列表中显示3个选项。这些是我数据库中的数据

例如：

公司a、公司b

将显示在下拉列表中

现在，下拉列表中只有一个选项

（公司a）

，而

（公司b）

显示在表格的下一行，而不是在单个下拉列表中一起显示

  <?
$result = mysqli_query($con,"SELECT admin_no, name, GPA, gender, job_details.job_title, job_details.no_of_vacancy FROM student_details, job_details ORDER BY `GPA` DESC ");

            $result2 = mysqli_query($con, "SELECT job_title FROM job_details;");
            $row2 = mysqli_fetch_assoc($result2);

            while($row = mysqli_fetch_assoc($result))
              {

                while ($row2 = mysqli_fetch_array($result2))
                  {


              echo "<tr>";
              echo "<td bgColor=white>" . $row['admin_no'] . "</td>";
              echo "<td bgColor=white>" . $row['name'] . "</td>";
              echo "<td bgColor=white>" . $row['GPA'] . "</td>";
              echo "<td bgColor=white>" . $row['gender'] . "</td>"; 
              echo "<td><select name='ddl' onclick='if(this.value != '') { myform.submit(); }'><option value='". $row2['job_title'] ."'> ". $row2['job_title'] ."</option></td>";
              echo "</tr>";
              }
              }
            echo "</table>";



    ?>
    </form>

试试这个：

while($row = mysqli_fetch_assoc($result))
          {

          echo "<tr>";
          echo "<td bgColor=white>" . $row['admin_no'] . "</td>";
          echo "<td bgColor=white>" . $row['name'] . "</td>";
          echo "<td bgColor=white>" . $row['GPA'] . "</td>";
          echo "<td bgColor=white>" . $row['gender'] . "</td>"; 

          echo "<td><select name='ddl' onclick='if(this.value != '') { myform.submit(); }'>";
       while ($row2 = mysqli_fetch_array($result2))
              {
         echo "<option value='". $row2['job_title'] ."'> ". $row2['job_title'] ."</option>";
            }
         echo "</select></td>";
          echo "</tr>";

          }

while（$row=mysqli\u fetch\u assoc（$result））
{
回声“；
回显“$行[“管理员编号]”；
回显“$row['name']”；
回显“$行['GPA']”；
回显“$row['gender']”；
回声“；
while（$row2=mysqli\u fetch\u数组（$result2））
{
回显“$row2['job_title']”；
}
回声“；
回声“；
}

试试这个，您没有关闭

标记，我已经重写了代码，即将

标记生成移到表生成部分上方。这样可以避免不必要的循环

    <?php
        $result = mysqli_query($con,"SELECT admin_no, name, GPA, gender, job_details.job_title, job_details.no_of_vacancy FROM student_details, job_details ORDER BY `GPA` DESC ");

            $result2 = mysqli_query($con, "SELECT job_title FROM job_details");
            $row2 = mysqli_fetch_assoc($result2);

             /*options sections start*/
            $options= '';
            while ($row2 = mysqli_fetch_array($result2))
            {
                $options .='<option value="'. $row2['job_title'] .'"> '. $row2['job_title'] .'</option>';
            }
            /*options sections end*/

            while($row = mysqli_fetch_assoc($result))
              {     
                  echo "<tr>";
                  echo "<td bgColor=white>" . $row['admin_no'] . "</td>";
                  echo "<td bgColor=white>" . $row['name'] . "</td>";
                  echo "<td bgColor=white>" . $row['GPA'] . "</td>";
                  echo "<td bgColor=white>" . $row['gender'] . "</td>"; 
                  echo "<td><select name='ddl' onclick='if(this.value != '') { myform.submit(); }'>".$options."</select></td>";
                  echo "</tr>";              
              }
            echo "</table>";
    ?>