Php 函数内的数据库查询未过帐到页面

我是php函数的新手，目前正在尝试构造一个if语句，该语句查看数据库表以查看名为tableformatting的列。此表存储了我的测验问题的格式选项，例如，如果我希望我的问题的答案显示为下拉菜单，我会在添加测验问题时选择该选项，然后当它从数据库中拉出时，它的格式将是这样的

目前，if语句如下所示：

    <?php
    // Lookup Question Format & Format Using Function //
    $questionformat="SELECT * FROM itsnb_chronoforms_data_createquestions WHERE questionID='$questionID'";
    $questionformatresult =mysql_query($questionformat);
    while ($formatrow = mysql_fetch_array($questionformatresult)){
    $format = $formatrow['questionformat']; }

            if ($format == 'questionformatdropmenu')
            {dropdownmenu ();
            }
            else
            {}

     ?> 

<?php
    // Question formatting functions //

    // Dropdown Menu
    function dropdownmenu(){

    echo '<select id="quizselectanswer" name="quizselectanswer" title="quizselectanswer">';

    $sql="SELECT * FROM itsnb_chronoforms_data_createquestions WHERE questionID='$questionID'";
    $result2 =mysql_query($sql);
    while ($data = mysql_fetch_array($result2)){

      echo '<option value ="' . $data['quizanswer1'] . '" >' . $data['quizanswer1'] . '</option>';
      echo '<option value ="' . $data['quizanswer2'] . '" >' . $data['quizanswer2'] . '</option>';
      echo '<option value ="' . $data['quizanswer3'] . '" >' . $data['quizanswer3'] . '</option>';
      echo '<option value ="' . $data['quizanswer4'] . '" >' . $data['quizanswer4'] . '</option>';
        }
      echo '</select>';
      }
    // Question formatting functions END //
    ?>

格式化函数如下所示：

    <?php
    // Lookup Question Format & Format Using Function //
    $questionformat="SELECT * FROM itsnb_chronoforms_data_createquestions WHERE questionID='$questionID'";
    $questionformatresult =mysql_query($questionformat);
    while ($formatrow = mysql_fetch_array($questionformatresult)){
    $format = $formatrow['questionformat']; }

            if ($format == 'questionformatdropmenu')
            {dropdownmenu ();
            }
            else
            {}

     ?> 

<?php
    // Question formatting functions //

    // Dropdown Menu
    function dropdownmenu(){

    echo '<select id="quizselectanswer" name="quizselectanswer" title="quizselectanswer">';

    $sql="SELECT * FROM itsnb_chronoforms_data_createquestions WHERE questionID='$questionID'";
    $result2 =mysql_query($sql);
    while ($data = mysql_fetch_array($result2)){

      echo '<option value ="' . $data['quizanswer1'] . '" >' . $data['quizanswer1'] . '</option>';
      echo '<option value ="' . $data['quizanswer2'] . '" >' . $data['quizanswer2'] . '</option>';
      echo '<option value ="' . $data['quizanswer3'] . '" >' . $data['quizanswer3'] . '</option>';
      echo '<option value ="' . $data['quizanswer4'] . '" >' . $data['quizanswer4'] . '</option>';
        }
      echo '</select>';
      }
    // Question formatting functions END //
    ?>

---

我遇到的问题是，函数创建下拉列表菜单很好，但是由于某种原因，数据没有被填充，下拉菜单只是空白，如果我将代码从函数中取出并正常运行，尽管它按预期工作，并且菜单中填充了答案。

您试图访问超出范围的

$questionID

变量。尝试将其传递给函数，例如：

// Dropdown Menu
function dropdownmenu ( $questionID ) {
  echo '<select id="quizselectanswer" name="quizselectanswer" title="quizselectanswer">';

  $sql="SELECT * FROM itsnb_chronoforms_data_createquestions WHERE questionID='$questionID'";
  $result2 =mysql_query($sql);

  while ( $data = mysql_fetch_array( $result2 ) ) {
    echo '<option value ="' . $data['quizanswer1'] . '" >' . $data['quizanswer1'] . '</option>';
    echo '<option value ="' . $data['quizanswer2'] . '" >' . $data['quizanswer2'] . '</option>';
    echo '<option value ="' . $data['quizanswer3'] . '" >' . $data['quizanswer3'] . '</option>';
    echo '<option value ="' . $data['quizanswer4'] . '" >' . $data['quizanswer4'] . '</option>';
  }

  echo '</select>';
}

---

您试图访问超出范围的

$questionID

变量。尝试将其传递给函数，例如：

// Dropdown Menu
function dropdownmenu ( $questionID ) {
  echo '<select id="quizselectanswer" name="quizselectanswer" title="quizselectanswer">';

  $sql="SELECT * FROM itsnb_chronoforms_data_createquestions WHERE questionID='$questionID'";
  $result2 =mysql_query($sql);

  while ( $data = mysql_fetch_array( $result2 ) ) {
    echo '<option value ="' . $data['quizanswer1'] . '" >' . $data['quizanswer1'] . '</option>';
    echo '<option value ="' . $data['quizanswer2'] . '" >' . $data['quizanswer2'] . '</option>';
    echo '<option value ="' . $data['quizanswer3'] . '" >' . $data['quizanswer3'] . '</option>';
    echo '<option value ="' . $data['quizanswer4'] . '" >' . $data['quizanswer4'] . '</option>';
  }

  echo '</select>';
}

---

谢谢你的帮助！，现在可以了：-），等时间允许的时候，我会马上标出答案。谢谢你的帮助！，现在可以了：-），我会在时间允许的时候标记答案。