javascript - PHP传递参数给新的字符串(学习)

Question

还在学习，还有很多问题，所以这里有一些。我正在进行 javascript -> PHP 转换，并希望确保这些做法是正确的。是$dailyparams->$calories = $calories;一条正确的线？再次感谢!

JavaScript

DailyParams.create4 = function(/*double*/ calories, /*double*/ carbpercent, /*double*/ sodium, /*double*/ actparam) {
if (calories < 0.0) calories = 0.0;
if (carbpercent < 0.0) carbpercent = 0.0;
if (carbpercent > 100.0) carbpercent = 100.0;
if (sodium < 0.0) sodium = 0.0;
if (actparam < 0.0) actparam = 0.0;

var dailyparams = new DailyParams();
dailyparams.calories = calories;
dailyparams.carbpercent = carbpercent;
dailyparams.sodium = sodium;
dailyparams.actparam = actparam;

return dailyparams;}
DailyParams.create2 = function(/*Intervention*/ inter, /*Baseline*/ base) {
var dailyparams = new DailyParams();
dailyparams.calories = inter.getcalories();
dailyparams.carbpercent = inter.getcarbinpercent();
dailyparams.sodium = inter.getsodium();
dailyparams.actparam = inter.getAct(base);

return dailyparams;

PHP

public function create4($calories, $carbpercent, $sodium, $actparam) {
    if ($calories < 0.0) $calories = 0.0;
    if ($carbpercent < 0.0) $carbpercent = 0.0;
    if ($carbpercent > 100.0) $carbpercent = 0.0;
    if ($sodium < 0.0) $sodium = 0.0;
    if ($actparam < 0.0) $actparam = 0.0;

    $dailyparams = new $DailyParams();
    $dailyparams->$calories = $calories;
    $dailyparams->$carbpercent = $carbpercent;
    $dailyparams->$sodium = $sodium;
    $dailyparams->$actparam = $actparam;

    return $dailyparams;
}
public function create2($inter,$base) {
    $dailyparams = new $DailyParams();
    $dailyparams->$calories = $inter->getcalories();
    $dailyparams->$carbpercent = $inter->getcarbinpercent();
    $dailyparams->$sodium = $inter->getsodium();
    $dailyparams->$actparam = $inter->getAct($base);

    return $dailyparams;
}

Answer 1

DailyParams 是类名，去掉$:

$dailyparams = new DailyParams();

与 calories 等属性名称相同:

$dailyparams->calories = $calories;