Question

美好的一天，我有一个从数据库中获取其值的表单，

<?php 
$questionId = 1;
  $pdo = new PDO("mysql:host=localhost;dbname=mydb", 'root', '');
    $sth = $pdo->prepare("SELECT * FROM questions WHERE questionid = $questionId");
    $sth->execute();

    while ($row = $sth->fetch()):
?>
  <form action="exam1.php" method="POST">
      <?php echo "<h4>".$row['instructions'] ."</h4><br>"; ?><h4><?php echo $questionId. ". ".$row['name']; ?></h4>
      <input type="radio" name="choices" id="choices" value="<?php echo $row['choice1']; ?>"><?php echo $row['choice1']; ?> ...(other choices)
      <input type="submit" name="submitAnswers">
  </form>
<?php endwhile;?>

我想要的是在提交表单后如何增加$questionId变量。我试过这个

if (isset($_GET['choices'])) {
    echo $_GET['choices'];
    $questionId = $questionId + 1;

  }

问题ID已更新，但只是一次。我需要它总是通过添加1来更新。

Answer 1

<form method="post" >

  <input type="submit" name="submit">
</form>
<?php 
session_start();
$_SESSION['id'];
if (isset($_POST['submit'])) {

    $_SESSION['id']=$_SESSION['id']+1;


}

echo "Value is => ",$_SESSION['id'],"<br>";
?>

PHP：增量错误 - 需要帮助

1 个答案: