php - 使用对象属性作为方法属性的默认值

Question

我正在尝试这样做（这会产生意外的 T_VARIABLE 错误）：

public function createShipment($startZip, $endZip, $weight = 
$this->getDefaultWeight()){}

我不想在其中输入一个幻数来表示重量，因为我使用的对象有一个"defaultWeight"参数，如果您不指定重量，所有新货物都会获得该参数。我不能将其defaultWeight放入货件本身，因为它会从货件组更改为货件组。有没有比以下更好的方法？

public function createShipment($startZip, $endZip, weight = 0){
    if($weight <= 0){
        $weight = $this->getDefaultWeight();
    }
}

Answer 1

这也好不到哪里去：

public function createShipment($startZip, $endZip, $weight=null){
    $weight = !$weight ? $this->getDefaultWeight() : $weight;
}

// or...

public function createShipment($startZip, $endZip, $weight=null){
    if ( !$weight )
        $weight = $this->getDefaultWeight();
}

Answer 2

布尔 OR 运算符的巧妙技巧：

public function createShipment($startZip, $endZip, $weight = 0){
    $weight or $weight = $this->getDefaultWeight();
    ...
}

Answer 3

这将允许您传递 0 的权重并且仍然可以正常工作。请注意 === 运算符，它检查权重是否在值和类型中都匹配“null”（与 == 相反，它只是值，因此 0 == null == false）。

PHP：

public function createShipment($startZip, $endZip, $weight=null){
    if ($weight === null)
        $weight = $this->getDefaultWeight();
}

Answer 4

您可以使用静态类成员来保存默认值：

class Shipment
{
    public static $DefaultWeight = '0';
    public function createShipment($startZip,$endZip,$weight=Shipment::DefaultWeight) {
        // your function
    }
}

Answer 5

如果您使用的是 PHP 7，请改进 Kevin 的回答，您可以这样做：

public function createShipment($startZip, $endZip, $weight=null){
    $weight = $weight ?: $this->getDefaultWeight();
}