PHP无法迭代太高或太低

[免责声明：我是PHP新手，我只是在学习，所以请别激动，这真的会阻碍学习过程，当你试图学习时，谢谢。 ]PHP无法迭代太高或太低

下面的代码运行，唯一的问题是它不告诉用户何时数字太高或太低，我做错了什么，但我看不到错误？

<?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>

来源

2009-11-07 Newb

我给+1的先发制人*请不要火焰我*这*真的*让我微笑=） – 2009-11-07 21:49:28

register_globals在php.ini可能是关（这是一件好事），因此您只能通过$_POST['Num']和$_POST['number']访问这些变量（你可以在你的脚本beggining）

也分配$number=$_POST['number']，发送的秘密$number通过形式是不好的，你可能想要阅读关于php sessions

来源

2009-11-08 00:27:26 jab11

$ number = $ _POST [ '号码']; $ Num = $ _POST ['Num']; 该死的register_globals，我将开始在我的所有表格上做这件事以避免麻烦，谢谢。 – Newb 2009-11-08 08:19:34

我假设$ Num是未定义的，我假设您假设它将被定义为因为它在窗体中定义。

在脚本的开始试试这个：

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

来源

2009-11-07 21:30:53

$number不会自动设置为<input>字段的值。（这是在早期版本的PHP中）。您现在必须为此使用$_POST['number']和$_POST['Num']。

来源

2009-11-07 21:33:51 Boldewyn

建议：

1）使用的回声，而不是echo

2）不要忘记关闭p标签

来源

2009-11-07 22:43:13 Joe

PHP无法迭代太高或太低

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