dph87312 2009-02-13 14:41 采纳率: 100%
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求助:猜数字游戏的问题

我的数字游戏就差一点就成功了,但出于某种原因我被卡住了。我将变量存储在一个隐藏的表单中,但是每次页面被发送时,它都会重置数字,然后就永远无法得到正确的数字。有什么想法吗?我的代码在下面:

<?php
// generate a random number for user to guess
$number = rand(1,100);

if($_POST["guess"]){

    // grab the user input guess
    $guess  = $_POST['guess'];
    $numbe  = $_POST['number'];
    if ($guess < $number){ 
        echo "Guess Higher";
    }elseif($guess > $number){       
        echo "Guess Lower";
    }elseif($guess == $number){      
        echo "You got it!";
    }
    echo "<br />Random Number:".$number."<br />";
    echo $guess;
}
?>

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8" />
<title>Guess A Number</title>
</head>

<body>
<form action="<?=$_SERVER['PHP_SELF'] ?>" method="post" name="guess-a-number">
    <label for="guess">Guess A Number:</label><br/ >
    <input type="text" name="guess" />
    <input name="number" type="hidden" value="<?= $number ?>" />
    <input name="submit" type="submit" />
</form>
</body>
</html>
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4条回答 默认 最新

  • douya1974 2009-02-13 14:45
    关注

    Change:

    // generate a random number for user to guess
    $number = rand(1,100);
    

    To:

    if(isset($_POST['number'])) {
       $number = $_POST['number'];
    } else {
       $number = rand(1,100);
    }
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论
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