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Home Backend Development PHP Tutorial 關于php獲取變量有關問題

關于php獲取變量有關問題

Jun 13, 2016 am 10:28 AM
amp keyboard quot

關于php獲取變量問題

PHP code
<!--Code highlighting produced by Actipro CodeHighlighter (freeware)http://www.CodeHighlighter.com/-->@$judge=$_GET["speed"];if(@$keyboard=$_GET["keyboard"]){$keyboardfinal=100;}if(@$judge==1){if($keyboard0){@$keyboardfinal=50;    }}if(@$judge==2){if(@$keyboard0){$keyboardfinal=50;    }}if(@$judge==3 && (@$keyboard0)){$keyboardfinal=50;    }

如上代碼~為什么在第二,三,四個if語句中,即使條件成立任然無法使$keyboardfinal=50;成立為什么呢?


------解決方案--------------------
@$judge=$_GET["speed"];
[email?protected]?
------解決方案--------------------
不明白樓主為什么要加那么多錯誤控制符,如果不報出錯誤,你怎么修改代碼呢?
將樓主的代碼修改了下:
PHP code
$judge = 1;$keyboard = 35;$keyboardfinal = null;switch($judge){    case 1:        if($keyboard0)            $keyboardfinal=50;        break;    case 2:        if($keyboard0)            $keyboardfinal=50;        break;    case 3 && ($keyboard0):        $keyboardfinal=50;        break;}echo "keyboardfinal-->>".$keyboardfinal;#50<br><font color="#e78608">------解決方案--------------------</font><br>
探討
PHP code
@$judge=$_GET["speed"];
if(@$keyboard=$_GET["keyboard"]){
$keyboardfinal=100;
}
if(@$judge==1){
if($keyboard0){
@$keyboardfinal=50;
}
}
if(@$judge==2){
if(@$keyb……
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