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将JavaScript代码转换为PHP_Php_Javascript - Fatal编程技术网

将JavaScript代码转换为PHP

将JavaScript代码转换为PHP,php,javascript,Php,Javascript,我需要帮助将此代码转换为PHP var formula = [10,6.7,3.4]; var user1 = [1,0,2,1,1,0,2,2,2,2]; var user2 = [2,1,2,0,1,2,2,2,0,0]; var result = 0; calculate(); function calculate(){ for(var i=0; i<user1.length; i++){ result +=formula[Math

我需要帮助将此代码转换为PHP

var formula = [10,6.7,3.4];  
var user1 = [1,0,2,1,1,0,2,2,2,2];  
var user2 = [2,1,2,0,1,2,2,2,0,0];  
var result = 0;  
calculate();  
function calculate(){  
    for(var i=0; i<user1.length; i++){  
        result +=formula[Math.abs(user1[i]-user2[i])]  
    }
}  
alert(result);
var公式=[10,6.7,3.4];
var user1=[1,0,2,1,1,0,2,2,2];
var user2=[2,1,2,0,1,2,2,2,0,0];
var结果=0;
计算();
函数calculate(){

对于(var i=0;i变量
$user1、$user2、$formula
不在范围内-将它们传递给函数:

function calculate( $user1, $user2, $formula) {  
    $result = 0;
    for( $i = 0, $size = count($user1); $i<$size; $i++){  
        $result += $formula[ abs( $user1[$i]-$user2[$i]) ]; 
    }
    return $result;
}  
$result = calculate( $user1, $user2, $formula);  
echo $result;

最后,我在返回计算值的函数中添加了
return$result;

必须传递变量并返回结果

$formula = array(10,6.7,3.4);  
$user1 = array(1,0,2,1,1,0,2,2,2,2);  
$user2 = array(2,1,2,0,1,2,2,2,0,0);  

function calculate($user1, $user2, $formula){  
    for($i=0, $size = count($user1); $i<$size; $i++){  
        $result += $formula[abs($user1[$i]-$user2[$i])];  
    }
    return $result;
}  
$result = calculate($user1, $user2, $formula);  
echo $result;  
$formula=数组(10,6.7,3.4);
$user1=数组(1,0,2,1,1,0,2,2,2);
$user2=数组(2,1,2,0,1,2,2,2,0,0);
函数计算($user1,$user2,$formula){

对于($i=0,$size=count($user1)$iIs有错误消息吗?另外,他在for循环中放弃了公式查找。您还应该
返回$result
。感谢各位的评论,我已经添加了您的建议并将
$result
初始化为
0
。您忘记了
返回值,我为您添加了它:-POh…Magic!:D(谢谢;)非常感谢大家
        $result += $formula[ abs( $user1[$i]-$user2[$i]) ];
$formula = array(10,6.7,3.4);  
$user1 = array(1,0,2,1,1,0,2,2,2,2);  
$user2 = array(2,1,2,0,1,2,2,2,0,0);  

function calculate($user1, $user2, $formula){  
    for($i=0, $size = count($user1); $i<$size; $i++){  
        $result += $formula[abs($user1[$i]-$user2[$i])];  
    }
    return $result;
}  
$result = calculate($user1, $user2, $formula);  
echo $result;