Question

我开发了验证码，要求用户回答随机显示的问题。我的数据库包括id，问题和答案。问题是虽然我输入了正确的答案，它仍然将我重定向到error.php而不是success.php。

＆＃13;

<?php

$database_db="test2";
$user_db="root";
$password_db="";
$host_db="localhost";

$link = mysqli_connect($host_db, $user_db, $password_db, $database_db);

/* check connection */
if (mysqli_connect_errno()) 
{
    die ("couldnot connect: ".mysqli_connect_error());
    exit();
}   
if (array_key_exists("answer", $_POST) AND array_key_exists("question", $_POST))
{
    $id = intval($_POST['question']);
    $sql="SELECT question, answer FROM captcha WHERE question='$id' AND              answer='".mysqli_real_escape_string($link, $_POST['answer'])."'";
    $result = mysqli_query($link, $sql)  or exit('$sql failed: '.mysqli_error($link)); 
    $num_rows = mysqli_num_rows($result);
    if($num_rows > 0)
    {
        header("Location: success.php");
    } 
    else 
    {
        header("Location: error.php");
    }
    exit;
}
else
{
    $query = "SELECT id, question FROM `captcha` ORDER BY RAND() LIMIT 1";
    if ($result = mysqli_query($link, $query))
    {
        if ($row = mysqli_fetch_assoc($result)) 
        {
            $id = $row["id"];
            $question = $row["question"];
        }
    }
 }

 ?>
<html>
<body>
    <form method="post">
        <?php echo $question; ?><br />
        <input type="hidden" name="question" id="question" value="<?php echo $id; ?>" />
        <input type="text" name="answer" id="answer" /><br />
        <input type="submit" name="submit" value="submit" /><br />
       </form>
    </body>
</html>

＆＃13;

Answer 1

您似乎在查询错误的列以查找与$id匹配的内容：

"SELECT question, answer FROM captcha WHERE question='$id' AND ... "

我认为这应该是：

"SELECT question, answer FROM captcha WHERE id='$id' AND ... "

如何验证数据库的输入

1 个答案: