Question

Question.php

<?php
include 'Pre-function.php'

?>
<!DOCTYPE html>
<html>
<head>
   <link rel="stylesheet" href="CSS/Start.css">
</head>
<body>

<div class="nav">
    <a href="Homepage.php">Home</a>
    <a href="#news">News</a>
    <a href="#contact">Contact</a>
</div> 

<div class="question"> 
    <div class="A4">
        <form action="Answer.php" method="post">
            <?php getQuestion($conn);  ?>
            <input type="submit" name="Submit" value="Submit">
        </form>
     </div>
 </div>
 </body>
 </html>

要查询问题的html页面

预function.php

<?php
include 'conn.php';

function getQuestion($conn) {

$query = "SELECT * FROM question ";
$result = mysqli_query($conn, $query);

if($result){

    while ($row = mysqli_fetch_array($result, MYSQLI_ASSOC)){
       $question_id = $row['question_id'];
       $question_body = $row['question_body'];
       $option_a = $row['option_a'];
       $option_b = $row['option_b'];

        echo '  
        <h2 class="qtitle">'.$question_body.'</h2>
        <label for='.$question_body.'>Yes</label>
            <input type="radio" name="'.$question_id.'" value="Yes">
            <input type="hidden" name="option_a" value="'.$option_a.'">

        <label for="'.$question_body.'">No</label>
            <input type="radio" name="'.$question_id.'" value="No">
            <input type="hidden" name="option_b" value="'.$option_b.'">

        <input type="hidden" name="submitted" value="submitted"><hr>';

        }

}

}

?>

基本上，此表单询问yes or no是否使用单选按钮。 $option_a == 'Yes'和$option_b == 'No'。问题是这样的＆＃34;你发烧了吗？＆＃34;。因此，当我提交的价值没有传递给＆＃39; Answer.php＆＃39;页。

＆＃39; Answer.php＆＃39;页。

<?php
include 'conn.php';

if(isset($_POST['Submit']) && !empty($_POST['Submit'])){

      echo $_POST['option_a'];
      echo 'succeed';
}
else{

     echo 'no data';    
}

?>

在此页面中有错误undefined_index值，但仍然是echo成功。

Answer 1

您的HTML代码应如下所示：

<input type="radio" name="whatevername" value="Yes">
<input type="hidden" name="whatevername" value="No">

您可以使用PHP插入所需的任何值，但您需要单选按钮名称相同。然后，如果你想在PHP中回应它，你可以使用：

echo $_POST['whatevername']; //same name you used in the form

Answer 2

您正在传递值作为单选按钮的名称

 name="'.$option_a.'"

这将导致您名字=“是”

您正在尝试提取echo $_POST['option_a'];，其中option_a未定义。

试试这个

<input type="radio" name="'.$question_id.'" value="Yes">
<input type="hidden" name="option_a" value="'.$option_a.'">

其他单选按钮

相同

Answer 3

    Try this one :

    <?php
include 'conn.php';

function getQuestion($conn) {

$query = "SELECT * FROM question ";
$result = mysqli_query($conn, $query);

if($result){
    echo '<div class="A4">';
    while ($row = mysqli_fetch_array($result, MYSQLI_ASSOC)){
        $question_id = $row['question_id'];
        $question_body = $row['question_body'];
        $option_a = $row['option_a'];
        $option_b = $row['option_b'];

        echo '
        <div class="A4">   
        <h2 class="qtitle">'.$question_body.'</h2>
        <form action="Answer.php" method="post">
            <label for='.$question_body.'>Yes</label>
            <input type="radio" name="radioQuestions[]" value="'.$question_id.'-'.$option_a.'">

            <label for="'.$question_body.'">No</label>
            <input type="radio" name="radioQuestions[]" value="'.$question_id.'-'.$option_b.'">

            <input type="hidden" name="submitted" value="submitted"><hr>';

    }
    echo'
        <input type="submit" name="Submit" value="Submit">
        </form>
        </div>';
}

}

?>

<?php
include 'conn.php';

if(isset($_POST['submitted']) && !empty($_POST['submitted'])){
     $questionAndOptions = $_POST['radioQuestions'];

     foreach ($questionAndOptions as $questionAndOption) {
        $arrQuestionAndOption =  explode("-", $questionAndOption);
         echo $arrQuestionAndOption[0]; //question
         echo $arrQuestionAndOption[1]; //option
     }


      echo 'succeed';
}
else{

     echo 'no data';    
}

?>

我如何从PHP中的单选按钮形式获取价值？

Question.php

预function.php

＆＃39; Answer.php＆＃39;页。

3 个答案: