php 如何避免否定答案

Question

嗨，我是新手，正在编写一个简单的税务代码。到目前为止，我已经写了一小段代码，但我认为函数 taxable_i($i) 似乎有问题，它不允许我生成 $taxable_i NIL。我已经尝试了一切。当我尝试通过函数制作 $taxable_i NIL 时，它不会让我这样做。它显示了 £13900 的结果。我认为该程序是在回应 $taxable_i 而不是应付税款。如果我将以上全部删除，那么应纳税额为 -£2199.80，为负。请注意，我输入 $taxable_i <personal_allowance 以便答案为零但结果不为零。我不知道为什么以及我做错了什么。欢迎任何帮助。'

<?php
// VAT 
define("vat_threshold",83000 );
define("vat_rate",0.2 );

// RELIEFS
define("personal_allowance",11000 );

//RATES
define("starting_rate",0 );
define("basic_rate",0.2 );
define("higher_rate",0.4 );
define("add_rate",0.45 );
define("divs_ord_rate",0.075 );
define("divs_uuper_rate",0.325 );
define("divs_add_rate",0.381 );

// THRESHOLDS
define("savings_income_band",5000 );
define("basic_rate_band",32000 );
define("higher_rate_band",150000 );
define("divs_allowance",5000 );

define ("band_0",0);
define("band_1",11000);
define("band_2",32000);
define("band_3",100000);
define("band_4",122000);
define("band_5",150000);

function taxable_i($i) {

    if ($i <= band_1 ) {
        $taxable_i = $i * 0;
        return $taxable_i;
    }
    if ($i <= band_3 ) {
        $taxable_i = $i - personal_allowance;
        return $taxable_i;
    } 
    if ($i >band_3 && $i <=band_4) {
        $taxable_i = $i - (personal_allowance-($i - band_3)/2);
        return $taxable_i;
    }
    if ($i > band_4) {
        $taxable_i = $i;
        return $taxable_i;
    }

}

$starting_income = $i = 1;
echo $i;    

$taxable_i = taxable_i($i);
echo $taxable_i;

switch ($taxable_i) {

case ($taxable_i > band_5):
    $diff = $taxable_i - band_5;
    $tax5 = $diff * add_rate;
    $taxable_i = band_5;
    $diff = $taxable_i - band_2;
    $tax4 = $diff * higher_rate;
    $taxable_i = band_2;
    $tax3 = $taxable_i * basic_rate;
    $tax_payable = $tax5 + $tax4 + $tax3;
    break;

case ($taxable_i > band_2 && $taxable_i <= band_5 ):
    $diff = $taxable_i - band_2;
    $tax4 = $diff * higher_rate;
    $taxable_i = band_2;
    $tax3 = $taxable_i * basic_rate;
    $tax_payable = $tax4 + $tax3;
    break;

case ($taxable_i < band_2):
    $tax = $taxable_i * basic_rate;
    $tax_payable = $tax;
    break;

default:    
    $taxable_i <= band_0;
    $tax_payable == 0;
    break;
}

echo $tax_payable;
?>

Answer 1

您不要在 case 语句中放置条件，您应该在那里使用 if/elseif/else。 switch/case 的工作方式是将初始 switch () 表达式的值与每个 case 表达式的值进行比较。所以它将 $taxable_i 与 $taxable_i > band_5 的值进行比较，因此它将一个数字与 true 或 false.

进行比较

将 switch 块替换为：

if ($taxable_i > band_5) {
    $diff = $taxable_i - band_5;
    $tax5 = $diff * add_rate;
    $taxable_i = band_5;
    $diff = $taxable_i - band_2;
    $tax4 = $diff * higher_rate;
    $taxable_i = band_2;
    $tax3 = $taxable_i * basic_rate;
    $tax_payable = $tax5 + $tax4 + $tax3;
}

elseif ($taxable_i > band_2 ) {
    $diff = $taxable_i - band_2;
    $tax4 = $diff * higher_rate;
    $taxable_i = band_2;
    $tax3 = $taxable_i * basic_rate;
    $tax_payable = $tax4 + $tax3;
}

elseif ($taxable_i >= band_0) {
    $tax = $taxable_i * basic_rate;
    $tax_payable = $tax;
}

else {    
    $tax_payable == 0;
    break;
}

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