Question

我正在尝试创建一个网页，提示用户猜测从1到5的数字，该数字随机生成并存储在表单中。

然后，当提交表单时，根据随机数检查猜测，如果猜测不正确，则提示用户再次猜测，但随机值与每次猜测保持相同。

我的问题是，一旦用户提交表单，当提示用户再次尝试时，随机数不会显示。

else if (isset($_POST["firstname"])) {
    echo $_POST['firstname'] . $_POST['lastname'];
    echo "Hi " . $_POST['firstname'] . " " . $_POST['lastname'] . "!";
    ?>
    <html>
        <form name="number" action="random.php" method="post">
            <p>Enter a guess: <input type="text" name="guess" /></p>
            <input type="hidden" name="numtobeguessed" value="<?php echo $_POST["numtobeguessed"]; ?>" ></p>
            <input type="submit" value="Guess!" />
        </form>
    </html>
    <?php
    $_POST["numtobeguessed"] = rand(1, 5);
    $guessed = $_POST["numtobeguessed"];
    echo "Number to be guessed " . $guessed;

} else if (isset($_POST["guess"])) {
    if ($_POST["guess"] != $_POST["guessed"]) {
        echo $_POST["guess"] . " is not correct";
        ?>
        <form name="number1" action="random.php" method="post">
            <p>Enter a guess: <input type="text" name="guess" /></p>
            <input type="hidden" name="numtobeguessed" value="<?php echo $_POST["guessed"]; ?>" ></p>
            <input type="submit" value="Guess!" />
            <?php
        }
    }
    ?>

Answer 1

您要在名称为numtobeguessed的字段中发送该值，但您尝试使用名称guessed进行阅读。

此：

if ($_POST["guess"] != $_POST["guessed"]) {

应该是：

if ($_POST["guess"] != $_POST["numtobeguessed"]) {

和此：

<input type="hidden" name="numtobeguessed" value="<?php echo $_POST["guessed"]; ?>" ></p>

应该是：

<input type="hidden" name="numtobeguessed" value="<?php echo $_POST["numtobeguessed"]; ?>" ></p>

PHP中的随机数生成器游戏使用表单

1 个答案: