函数

Question

我有以下PHP代码：

$car1 = new Car('Ford','Fusion');
$car2 = new Car('Chevy', 'Avalanche');
$car3 = new Car('Ford', 'F150');

$cars = array($car1, $car2, $car3);

function getCarsByMake($carMake){
    foreach($cars as $car){
        if($car->make == $carMake){
            echo 'Car: ' . $car->make . ' ' . $car->model . "<br>";
        }
    }
}

getCarsByMake('Ford');

我收到$cars语句中foreach未定义的错误。但是，据我所知，$cars数组的范围应该是全局的？如果我通过构造函数将数组传递给函数，它可以正常工作。但我想知道为什么我不能以这种方式访问​​阵列。

Answer 1

除了Exprator的解决方案，您还可以将$cars数组传递给这样的函数。

function getCarsByMake($carMake, $cars){
    foreach($cars as $car){
        if($car->make == $carMake){
            echo 'Car: ' . $car->make . ' ' . $car->model . "<br>";
        }
    }
}

getCarsByMake('Ford', $cars);

Answer 2

function getCarsByMake($carMake){
    global $cars;
    foreach($cars as $car){
        if($car->make == $carMake){
            echo 'Car: ' . $car->make . ' ' . $car->model . "<br>";
        }
    }
}

getCarsByMake('Ford');

因为函数没有得到$cars，你需要在函数内全局访问它

Answer 3

您有两种选择。

1. 添加global关键字

   function getCarsByMake($carMake){
      global $cars;
      foreach($cars as $car){
          if($car->make == $carMake){
              echo 'Car: ' . $car->make . ' ' . $car->model . "<br>";
          }
      }
   }
   

2. 使用$GLOBALS数组：

   function getCarsByMake($carMake){
       foreach($GLOBALS["cars"] as $car){
           if($car->make == $carMake){
               echo 'Car: ' . $car->make . ' ' . $car->model . "<br>";
           }
       }
   }
   

虽然我仍然建议将其作为显式参数传递，因为这样可以使代码更具可读性和可维护性，恕我直言。

Answer 4

您可以在全球范围内拨打$ cars，但您也可以根据您的代码尝试其他方式.Ex：

$car1 = new Car('Ford','Fusion');
$car2 = new Car('Chevy', 'Avalanche');
$car3 = new Car('Ford', 'F150');

$cars = array($car1, $car2, $car3);

function getCarsByMake($carMake,$cars){
    foreach($cars as $car){
        if($car->make == $carMake){
            echo 'Car: ' . $car->make . ' ' . $car->model . "<br>";
        }
    }
}

getCarsByMake('Ford',$cars);