Question

下面的PHP脚本适用于数字猜谜游戏。隐藏字段用于存储猜测的数量，由变量$ tries表示。无论我如何尝试，猜测（$尝试）的数量永远不会增加。有人能告诉我代码有什么问题吗？玛利亚

<?php
$num=21;
$tries=(isset($_POST['guess'])) ? $tries+1: 0;
if (!isset($_POST['guess'])) {
    $message="Welcome to the Guessing Game!";
} elseif (!is_numeric($_POST['guess'])) {
    $message="You need to type in a number.";
} elseif ($_POST['guess']==$num) {
    $message="You WON 1 million point!";
} elseif ($_POST['guess']>$num) {
    $message="Try a smaller number";
} else {
    $message="Try a bigger number";
}
?>
<!DOCTYPE html>
<html>
<head>
<title>Guessing Game</title>
</head>
<body>
<h1><?php echo $message; ?></h1>
<p><strong>Guess number: </strong><?php echo $tries; ?></p>
<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="POST">
<p><label for="guess">Type your guess here:</label><br/>
<input type="text" id="guess" name="guess" />
<input type="hidden" name="tries" value="<?php echo $tries; ?>"/></p>
<button type="submit" name="submit" value="submit">Submit</button>
</form>
</body>
</html>

Answer 1

变化：

$tries=(isset($_POST['guess'])) ? $tries+1: 0;

要：

$tries=(isset($_POST['tries'])) ? $_POST['tries']+1: 0;

或者，根据条件，仅在猜测存在时递增，然后：

$tries=(isset($_POST['guess'])) ? $_POST['tries']+1: 0;

但是基于guess进行此操作将允许此人通过不输入数字来重置计数器;）

Answer 2

当您尝试递增时，未定义

$tries。但是它可以在$_POST

中使用

$tries=(isset($_POST['guess'])) ? $tries+1: 0;
// Change to
$tries=(isset($_POST['guess'])) ? $_POST['tries']+1: 0;

在PHP猜数游戏中，尝试次数不会增加

2 个答案: