Php 如何在同一数组中引用变量 $price=$product->price; $priceData=[ “小计”=>美元价格, “税”=>$price*0.13, “总计”=>$price+>>>>$tax
不是您尝试的方式。您有三种选择:Php 如何在同一数组中引用变量 $price=$product->price; $priceData=[ “小计”=>美元价格, “税”=>$price*0.13, “总计”=>$price+>>>>$tax,php,arrays,variables,Php,Arrays,Variables,不是您尝试的方式。您有三种选择: $price = $product->price; $priceData = [ 'subtotal' => $price, 'tax' => $price * 0.13, 'total' => $price + >>>> $tax <<<< ]; 或: 或: 您需要将计算与结构分离,并在计算后创建结构 $tax = $price * 1.13; $priceDat
$price = $product->price;
$priceData = [
'subtotal' => $price,
'tax' => $price * 0.13,
'total' => $price + >>>> $tax <<<<
];
或:
或:
您需要将计算与结构分离,并在计算后创建结构
$tax = $price * 1.13;
$priceData = [
'subtotal' => $price,
'tax' => $tax,
'total' => $price + $tax
];
$price
数组是否应该将自身作为一个值包含?不,我在发布之前从项目中更改了此代码。我将更改数组的名称以避免混淆。是否有原因不能首先在数组之外执行$price+$tax计算?$tax
未定义,但$priceData['tax']
is$priceData['tax']不起作用
$priceData = [
'subtotal' => $price,
'tax' => $price * 1.13
];
$priceData['total'] = $price + $priceData['tax'];
$tax = $price * 1.13;
$priceData = [
'subtotal' => $price,
'tax' => $tax,
'total' => $price + $tax
];
$tax = $price * 1.13;
$total = $price + $tax;
$data = [
'subtotal' => $price,
'tax' => $tax,
'total' => $total
];