表格中的下拉列表答案

【问题标题】：Drop down list in a table表格中的下拉列表
【发布时间】：2014-01-24 09:44:58
【问题描述】：

为什么我的下拉列表在表格中分开来区分列。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： company a, company b 将显示在下拉列表中。

现在下拉列表中只有一个选项 (company a) 可用，(company b) 显示在表格的下一行，而不是一起显示在单个下拉列表中。

  <?
$result = mysqli_query($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>

【问题讨论】：

标签： php html-table

【解决方案1】：

试试这个，你错过了关闭</select>标签，我已经重写了代码，即将<option>标签生成移到表格生成部分上方。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>";
    ?>

【讨论】：

【解决方案2】：

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>";

          }

【讨论】：