Question

我正在尝试通过检查mysqli变量来编译if语句以检查回复类型是Single还是Multiple。如果Single然后输出选项作为单选按钮，如果Multiple则输出选项作为复选框。

但我在$qandaReplyType的if语句中收到一个未定义的变量错误。如何解决这个问题？

以下代码：

    $qandaquery = "SELECT q.QuestionId, r.ReplyType
                    FROM Question q
                    LEFT JOIN Reply r ON q.ReplyId = r.ReplyId

                    WHERE SessionId = ?
                    GROUP BY q.QuestionId
                    ORDER BY RAND()";

    $qandaqrystmt=$mysqli->prepare($qandaquery);
    // get result and assign variables (prefix with db)
    $qandaqrystmt->execute(); 
    $qandaqrystmt->bind_result($qandaQuestionId,$qandaReplyType);


    $arrReplyType = array();


    while ($qandaqrystmt->fetch()) {

    $arrReplyType[ $qandaQuestionId ] = $qandaReplyType;
  }

    $qandaqrystmt->close();





    function ExpandOptionType($option) { 

    $options = explode('-', $option);
    if(count($options) > 1) {
        $start = array_shift($options);
        $end = array_shift($options);
        do {
            $options[] = $start;
        }while(++$start <= $end);
     }
     else{
        $options = explode(' or ', $option);
     }

     if($qandaReplyType == 'Single'){
     foreach($options as $indivOption) {
         echo '<div class="ck-button"><label class="fixedLabelCheckbox"><input type="radio" name="options_<?php echo $key; ?>[]" id="option-' . $indivOption . '" value="' . $indivOption . '" /><span>' . $indivOption . '</span></label></div>';
     }
 }else if($qandaReplyType == 'Multiple'){
           foreach($options as $indivOption) {
         echo '<div class="ck-button"><label class="fixedLabelCheckbox"><input type="checkbox" name="options_<?php echo $key; ?>[]" id="option-' . $indivOption . '" value="' . $indivOption . '" /><span>' . $indivOption . '</span></label></div>';
     }
}
}

foreach ($arrQuestionId as $key=>$question) {

echo ExpandOptionType(htmlspecialchars($arrOptionType[$key]));

}

?>

Answer 1

函数内部不存在

$qandaReplyType。

在使用变量的全局版本之前，您需要在函数中包含global $qandaReplyType。

在PHP中使用variable scope时不会受到伤害。

修改：根据它的外观，$arrQuestionId和$arrOptionType也不在功能范围内。
global $qandaReplyType, $arrQuestionId, $arrOptionType;

Answer 2

Mmmmmmm .....你的功能......试试这个：

function ExpandOptionType($option) { 
$qandaReplyType = "Single";
$options = explode('-', $option);
if(count($options) > 1) {
    $start = array_shift($options);
    $end = array_shift($options);
    do {
        $options[] = $start;
    }while(++$start <= $end);
 }
 else{
    $options = explode(' or ', $option);
 }

 if($qandaReplyType == 'Single'){
 foreach($options as $indivOption) {
     echo '<div class="ck-button"><label class="fixedLabelCheckbox"><input type="radio"
    name="options_<?php echo $key; ?>[]" id="option-' . $indivOption . '" value="' . 
 $indivOption . '" /><span>' . $indivOption . '</span></label></div>';
 }
 }else if($qandaReplyType == 'Multiple'){
       foreach($options as $indivOption) {
     echo '<div class="ck-button"><label class="fixedLabelCheckbox"><input type="checkbox" name="options_<?php echo $key; ?>[]" id="option-' . $indivOption . '" value="' . $indivOption . '" /><span>' . $indivOption . '</span></label></div>';
 }
 }
 }

如果这样可行，那么$ qandaReplyType没有在函数内部定义....你应该找到一种方法来恢复该值o可能会在函数中添加另一个参数并传递$ qandaReplyType值，如：

function ExpandOptionType($option,$qandaReplyType)

Saludos;）

if语句中的未定义变量

2 个答案: