Question

[免责声明：我是PHP的新手，我只是在学习，所以请不要浪费，当人们试图学习时，它确实会阻碍学习过程，谢谢。]

下面的代码运行，唯一的问题是它没有告诉用户什么时候数字太高或太低，我做错了什么，但我看不出错误？

<?php
//Starts our php document
if (!$number)
//if we have already defined number and started the game, this does not run
{
 Echo"Please Choose a Number 1-100 <p>";
//gives the user instructions
$number = rand (1,100) ;
//creates number
}

else {
//this runs if the game is already in progress
if ($Num >$number)
{
Echo "Your number, $Num, is too high. Please try again<p>";
}
//if the number they guessed is bigger than number,  lets user know,  guess was  high

elseif ($Num == $number)
{
Echo "Congratulations you have won!<p>";
//if the number they guessed was correct it lets them know they won
Echo "To play again, please Choose a Number 1-100 <p>";
$number = rand (1,100) ;
//it then starts the game again by choosing a new value for $number that they can guess
}

else
{
Echo "Your number, $Num, is too low. Please try again<p>";
}
//if the answer is neither correct or to high, it tells them it is too low
}
?> 

<form action = "<?php Echo $_SERVER[’PHP_SELF’]; ?>" method = "post"> <p>
<!--this sends the form back to the same page we are on-->

Your Guess:<input name="Num" />
<input type = "submit" name = "Guess"/> <p>
<!--Allows the user to input their guess-->

<input type = "hidden" name = "number" value=<?php Echo $number ?>>
<!--keeps passing along the number value to keep it consistent till it is guessed-->

</form>
</body>
</html>

Answer 1

我假设$ Num未定义，我假设您假设它将被定义为因为它在表单中定义。

在脚本开头尝试：

if(!empty($_POST)) {
    $Num = (int) $_POST['Num'];
}

Answer 2

$number不会自动设置为<input>字段的值。（它是在PHP的早期版本中）。您现在必须使用$_POST['number']和$_POST['Num']。

Answer 3

你的php.ini中的

register_globals可能是Off（这是一件好事），因此你只能通过$_POST['Num']和$_POST['number']访问这些变量（你可以只分配{{} 1}}在你的剧本开始时）

另外，通过表单发送秘密$number=$_POST['number']并不好，您可能想了解php sessions

Answer 4

建议：

1）使用echo，而不是Echo

2）不要忘记关闭p标签

PHP无法迭代太高或太低

4 个答案: