使用下一步按钮替换我页面上的内容，而不将我重定向到新页面

Question

我正在尝试制作一个测验应用程序，它在两侧显示带有单选按钮的答案。当您按下下一步按钮时，将出现一组新答案并替换 .

我已经设法按预期弹出四个问题，并在我按下下一步按钮时弹出四个新问题。现在有一个问题，我的第一组四个答案（qid = 1）没有消失，这很奇怪，因为另一组 qid = 2 和 3 的答案会在我按下一个按钮时互相替换。

如何才能让新答案出现并替换旧答案？

到目前为止，这是我的代码 PHP:

$qid1 = 1;

$sql1 = mysqli_query($connect,"SELECT * FROM question where qid ='$qid1'");

while($row=mysqli_fetch_assoc($sql1))
{

echo "<input type='radio' name='answer1' value='".$row['Point']."'>"
        .$row['answer'] ."<br>";

}
echo "<input type='submit' name='forward1' value='next'>";
$qid2 = 2;

$sql2 = mysqli_query($connect,"SELECT * FROM question where qid ='$qid2'");


while($row2=mysqli_fetch_assoc($sql2)){
if (isset($_POST['forward1'])) {


echo "<input type='radio' name='answer2' value='".$row2['Point']."'>"
        .$row2['answer'] ."<br>";

}
}

echo "<input type='submit' name='forward2' value='next'>";
$qid3 = 3;

$sql3 = mysqli_query($connect,"SELECT * FROM question where qid ='$qid3'");

while($row3=mysqli_fetch_assoc($sql3)){
if (isset($_POST['forward2'])) {


echo "<input type='radio' name='answer3' value='".$row3['Point']."'>"
        .$row3['answer'] ."<br>";

}
}
echo "<input type='submit' name='forward3' value='next'>";

Answer 1

您需要使用表单标签分隔您的输入。对于你的每个循环做这样的事情..

echo "<form>";
$sql3 = mysqli_query($connect,"SELECT * FROM question where qid ='$qid3'");
while($row3=mysqli_fetch_assoc($sql3)){
    if (isset($_POST['forward2'])) {
        echo "<input type='radio' name='answer3' value='".$row3['Point']."'>"
        .$row3['answer'] ."<br>";
    }
}
echo "<input type='submit' name='forward3' value='next'>";
echo "</form>";

试试这个..

<?php

$localhost = "localhost";
$username = "root";
$password = "";
$connect = mysqli_connect($localhost, $username, $password) || die("Kunde inte koppla");
mysqli_select_db($connect, 'wildfire');

// let's put the qid in a session var
session_start();
$qid = isset($_SESSION['qid']) ? $_SESSION['qid']+1 : 1;
$_SESSION['qid'] = $qid;

ob_start();

echo "<form>";
$sql1 = mysqli_query($connect,"SELECT * FROM question where qid ='$qid'");

while($row1=mysqli_fetch_assoc($sql1))
    echo "<input type='radio' name='answer1' value='{$row1['Point']}'>{$row1['answer']}<br>";

echo "<input type='submit' name='forward1' value='next'>";
echo "</form>";


$output = ob_get_clean();


?>

?><!DOCTYPE html>
<html>
    <head>
        <meta charset="utf-8">
    </head>
    <body> 
        <?php echo $output; ?>
    </body>
</html>

使用下一步按钮替换我页面上的内容，而不将我重定向到新页面

Using the next button to replace content on my page, without redirecting me to a new page

php

mysql

button

radio-button

next