使用对象属性作为方法属性的默认值

时间:2020-03-05 18:37:35  来源:igfitidea点击:

我正在尝试执行此操作(这会产生意外的T_VARIABLE错误):

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

我不想在其中放一个魔术数字来表示重量,因为我使用的对象有一个" defaultWeight"参数,如果我们未指定重量,所有新货都会得到该参数。我不能将defaultWeight放入货件本身,因为它在货件组之间变化。有没有比以下更好的方法了?

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

解决方案

回答

这不是更好:

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();
}

回答

这将使我们传递的权重为0并仍然可以正常工作。注意===运算符,它检查权重在值和类型上是否都匹配" null"(与==相反,后者只是值,所以0 == null == false)。

PHP:

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

回答

我们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
    }
}

回答

布尔OR运算符的巧妙窍门:

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