关于php获取变量有关问题_PHP

关于php获取变量问题

PHP code

@$judge=$_GET["speed"];if(@$keyboard=$_GET["keyboard"]){$keyboardfinal=100;}if(@$judge==1){if($keyboard<40 && $keyboard>0){@$keyboardfinal=50;    }}if(@$judge==2){if(@$keyboard<50&& $keyboard>0){$keyboardfinal=50;    }}if(@$judge==3 && (@$keyboard<170 && @$keyboard>0)){$keyboardfinal=50;    }

如上代码~为什么在第二，三，四个if语句中，即使条件成立任然无法使$keyboardfinal=50;成立为什么呢？

------解决方案--------------------
@$judge=$_GET["speed"];
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------解决方案--------------------
不明白楼主为什么要加那么多错误控制符，如果不报出错误，你怎么修改代码呢？
将楼主的代码修改了下：

PHP code

$judge = 1;$keyboard = 35;$keyboardfinal = null;switch($judge){    case 1:        if($keyboard<40 && $keyboard>0)            $keyboardfinal=50;        break;    case 2:        if($keyboard<50&& $keyboard>0)            $keyboardfinal=50;        break;    case 3 && ($keyboard<170 && $keyboard>0):        $keyboardfinal=50;        break;}echo "keyboardfinal-->>".$keyboardfinal;#50
------解决方案--------------------
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PHP code
@$judge=$_GET["speed"];
if(@$keyboard=$_GET["keyboard"]){
$keyboardfinal=100;
}
if(@$judge==1){
if($keyboard<40 &amp;&amp; $keyboard>0){
@$keyboardfinal=50;
    }
}
if(@$judge==2){
if(@$keyb……