使用下一个按钮替换页面上的内容，而无需将我重定向到新页面

问题描述：

I'm trying to make a quiz application that shows answers with radio buttons on the sides. When you press the next button a set of new answers will appear and replace the .

I've managed to make four questions pop up as intended and four new ones popup when I press the next button. Right now theres one problem, my first set of four answers (with qid = 1) does not dissapear, which is weird since the other set of answers with qid = 2 and 3 does and replace eachother whenever I press the next button.

How do I make it so that the new answers appear and replace the old answers?

Here is my code so far 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'>";

我正在尝试制作一个测验应用程序，用两侧的单选按钮显示答案。当你按下一个按钮时，会出现一组新的答案并替换。 p>

我设法按预期弹出四个问题，当我按下一个时弹出四个新问题按钮。现在有一个问题，我的第一组四个答案（qid = 1）不会消失，这很奇怪，因为qid = 2和3的另一组答案确实并且每当我按下一个按钮时替换另一个。 p>

如何制作新答案并替换旧答案？ p>

这是我的代码到目前为止PHP： p>

  $ qid1 = 1; 
 
 $ sql1 = mysqli_query（$ connect，“SELECT * FROM question where qid ='$ qid1'”）; 
 
而（$  row = mysqli_fetch_assoc（$ sql1））
 {
 
echo“＆lt; input type ='radio'name ='answer1'value ='”。$ row ['Point']。“'＆gt;”
。  $ row ['answer']。“＆lt; br＆gt;”; 
 
} 
echo“＆lt; input type ='submit'name ='forward1'value ='next'＆gt;”; 
 $ qid2 =  2; 
 
 $ sql2 = mysqli_query（$ connect，“SELECT * FROM question where qid ='$ qid2'”）; 
 
 
而（$ row2 = mysq  li_fetch_assoc（$ sql2））{
if（isset（$ _ POST ['forward1']））{
 
 
echo“＆lt; input type ='radio'name ='answer2'value ='”。$ row2 [  'Point']。“'＆gt;”
。$ row2 ['answer']。“＆lt; br＆gt;”; 
 
} 
} 
 
echo“＆lt; input type ='submit'name  ='forward2'value ='next'＆gt;“; 
 $ qid3 = 3; 
 
 $ sql3 = mysqli_query（$ connect，”SELECT * FROM question where qid ='$ qid3'“）; 
 \  nwhile（$ row3 = mysqli_fetch_assoc（$ sql3））{
if（isset（$ _ POST ['forward2']））{
 
 
echo“＆lt; input type ='radio'name ='answer3'value ='  “。$ row3 ['Point']。”'＆gt;“
。$ row3 ['answer']。”＆lt; br＆gt;“; 
 
} 
} 
echo”＆lt; input type =' 提交'name ='forward3'值='下一个'＆gt;“; 
  code>  pre> 
  div>

答

You need to separate your inputs with form tags. For each of your loops do something like this..

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>";

Try this..

<?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>

使用下一个按钮替换页面上的内容，而无需将我重定向到新页面

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