PHP的基本计算器的功能和开关，但结果是

Question

我目前正在为练习做一些简单的计算器，但此处未显示输出或结果，这是我的代码人员希望您能为我提供帮助：/

    <input type="radio" value= "Addition" name="calcu"> Addition .<br />
    <input type="radio" value= "Subtraction" name="calcu"> Subtraction .<br />
    <input type="radio" value= "Multiplication" name="calcu"> Multiplication .<br />
    <input type="radio" value= "Division" name="calcu"> Division .<br />
<?php
$num1 = $_POST['num1'];
$num2 = $_POST['num2'];
$calcu = $_POST['calcu'];

    function calculate($n1,$n2)
    {
        switch('$calcu')
        {
        case "Addition";
            $compute = $n1 + $n2; 
            break;
        case "Subtraction";
            $compute = $n1 - $n2; 
            break;
        case "Multiplication";
            $compute = $n1 * $n2; 
            break;
        case "Division";
            $compute = $n1 / $n2; 
            break;
        }
    }
echo "$calcu <br /> <br /> 1st Number: $num1 <br /> 2nd Number: $num2 <br /><br />";
echo "Answer is:" .calculate($num1,$num2);
?>

Answer 1

这是完整的代码：

<?php
$num1 = $_POST['num1'];
$num2 = $_POST['num2'];
$calcu = $_POST['calcu'];

    function calculate($n1,$n2, $calcu) // set $calcu as parameter
    {
        switch($calcu)
        {
        case "Addition": // here you have to use colons not semi-colons
            $compute = $n1 + $n2; 
            break;
        case "Subtraction":
            $compute = $n1 - $n2; 
            break;
        case "Multiplication":
            $compute = $n1 * $n2; 
            break;
        case "Division":
            $compute = $n1 / $n2; 
            break;
        }
        return $compute; // returning variable
    }
echo "$calcu <br /> <br /> 1st Number: $num1 <br /> 2nd Number: $num2 <br /><br />";
echo "Answer is:" .calculate($num1,$num2, $calcu); // you need to pass $calcu as argument of that function
?>

Answer 2

将switch('$calcu')更改为switch($calcu) 。

如@PeterM所述，您正在访问变量$calcu超出范围。 您可以传递$calcu变量进行有趣的calculate也可以直接通过$_POST数组进行访问。

使用switch($_POST['calcu']) 。

要么

function calculate($n1,$n2, $calcu) {
...
}

通过calculate($n1,$n2, $calcu)调用乐趣。

Answer 3

将switch('$calcu')更改为switch($calcu) 。 应该是这种方式。

但不仅如此。 您的变量未定义，因为您正在尝试在提交表单之前解决它们，即它们尚不存在。

$num1 = $_POST['num1'];
$num2 = $_POST['num2'];
$calcu = $_POST['calcu'];

在那儿你解决他们

echo "$calcu <br /> <br /> 1st Number: $num1 <br /> 2nd Number: $num2 <br /><br />";
echo "Answer is:" .calculate($num1,$num2);

实现此目的的正确方法是检查表单是否已提交：

    <input type="radio" value= "Addition" name="calcu"> Addition .<br />
    <input type="radio" value= "Subtraction" name="calcu"> Subtraction .<br />
    <input type="radio" value= "Multiplication" name="calcu"> Multiplication .<br />
    <input type="radio" value= "Division" name="calcu"> Division .<br />
<?php
if (isset($_POST)){
    $num1 = $_POST['num1'];
    $num2 = $_POST['num2'];
    $calcu = $_POST['calcu'];

        function calculate($n1,$n2)
        {
            switch('$calcu')
            {
            case "Addition";
                $compute = $n1 + $n2; 
                break;
            case "Subtraction";
                $compute = $n1 - $n2; 
                break;
            case "Multiplication";
                $compute = $n1 * $n2; 
                break;
            case "Division";
                $compute = $n1 / $n2; 
                break;
            }
        }
    echo "$calcu <br /> <br /> 1st Number: $num1 <br /> 2nd Number: $num2 <br /><br />";
    echo "Answer is:" .calculate($num1,$num2);

    unset($_POST);
}
?>