Why is it always an error, even when entering the correct number?

B

BlackTritium2014-03-02 08:43:45

PHP

BlackTritium, 2014-03-02 08:43:45

<html>
<head>
    <meta charset="utf-8">
    <title>Bender Rodriguez v.0.1</title>
</head>
<body>
<h1>Bender v.0.1</h1>
<p>Сыграем?</p>
<img id="Bender" src="http://diabloprint.ru/img/Bender_Rodriguez.png" alt="Bender"><br>
<?php
$numOne = rand(1,10);
$numTwo = rand(1,10);
$answer = $numOne + $numTwo;
echo ('Ответ:');
echo($answer);
echo ('<br>Сколько будет:');
echo ($numOne);
echo (' + ');
echo ($numTwo);
echo ('<br>Мой ответ:');
echo htmlspecialchars($_POST["answerUser"]);
if($answerUser===$answer) // or if(call())
{
    echo('<br>Правильно');
}
else
{
    echo('<br>Неправильно');
}

$answered = $answer;
?>
<form method="post" action="index.php">
    <input name="answerUser" type="text" maxlength="20" size="25" value="" />
    <input type=submit value="Ответить"></form>

</body>
</html>

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2 answer(s)

I

Ivan Starkov, 2014-03-02
@icelaba

$answerUser===$answer must match types, most likely in this scenario answer is of type Number and answerUser is String
so either use == or explicitly cast to Number of type
1*$answerUser === $answer

A

Alexander Zelenin, 2014-03-02
@zelenin

htmlspecialchars($_POST["answerUser"]) most likely casts a number to a string
if($answerUser===$answer) - three = this is equal, given the type (integer or string).
Either the string must be converted to an integer or replaced with ==. Better first.