Question

我遇到一个小问题，我想在数据库中检查问题的答案，

我正在根据传入的问题ID进行查询并将答案设置为变量，然后我想将变量与用户输入进行比较;我犯了一个愚蠢的错误，但无法理解。

更新＃1：
无论输入什么，它只是显示错误我也尝试将它们与！=

进行比较

<?php

// establishing the MySQLi connection

require_once('connection-test.php');
$mysqli = new mysqli($host,$username,$password,$database);

    if($mysqli -> connect_error)die($mysqli->connect_error);


$questionID = $_POST['id'];

$dbAnswer = "SELECT answers FROM answers where questionID='$questionID'";

$answer = mysqli_real_escape_string($mysqli,$_POST['answer']);
if ($result3 = $mysqli->query($dbAnswer)) {

    /* fetch object array */
    while ($obj = $result3->fetch_object()) {
        //printf ("%s  \n", $obj->answers);
        $ans = $obj->answers;

        echo $ans;

    }
    if($ans != $result3){
        echo "<b>error</b>" . $answer;
    }else{
         echo "<br><br><br>";
        echo "Correct" . $answer;
    }

    /* free result set */
    $result3->close();
}

?>

Answer 1

我已修复了我的问题，现在一切正常，这是我使用的代码

// establishing the MySQLi connection

require_once('connection-test.php');
$mysqli = new mysqli($host,$username,$password,$database);

if($mysqli -> connect_error)die($mysqli->connect_error);


$questionID = $_POST['id'];
$userAnswer = $_POST['answer'];

$query = "SELECT answers FROM answers WHERE questionID=?";
$statement = $mysqli->prepare($query);
$statement ->bind_param('i', $questionID);

$statement->execute();

$statement->bind_result($answer);


while($statement->fetch()){
    if($answer != $userAnswer){
        echo "Wrong";
    }else{
        echo "correct!";
}

}

 $statement->close();

比较用户输入到数据库

1 个答案: