在PHP中按类（instanceof）切换

Question

可以用开关代替

if( .. instanceof ...), elseif(... instanceof ...), ...

的块吗？

例如：

<?php
$class = ..... //some class

if($class instanceof SomeClass) {
    //do something
} elseif($class instanceof SomeAnotherClass) {
    //do something else
}

Answer 1

对于带有

instanceof

且考虑继承的多态开关：

switch(true) {  
    case $objectToTest instanceof TreeRequest:
        echo "tree request";
        break;
    case $objectToTest instanceof GroundRequest:
        echo "ground request";
        break;
}

对于类名应完全匹配的开关：

$class = get_class($objectToTest);

switch($class) {  
    case 'TreeRequest':
        echo "tree request";
        break;
    case 'GroundRequest':
        echo "ground request";
        break;
}

Answer 2

以下说法更正确：

$class = get_class($objectToTest);

switch($class) {  
    case TreeRequest::class:
        echo "tree request";
        break;
    case GroundRequest::class:
        echo "ground request";
        break;
}

通过这种方式，可以正确比较命名空间（您不必将它们键入），如果命名空间或类名等发生任何更改，您将收到一个不错的 IDE 错误，通知您代码已损坏。

Answer 3

使用

match

表达式，看起来像这样：

return match(get_class($object)) {  
   TreeRequest::class => "tree request",
   GroundRequest::class => "ground request",
   default => "default request"
};

记住它缺少继承并且仅采用当前类。考虑继承的匹配表达式：

return match(true) {  
   $object instanceof TreeRequest=> "tree request",
   $object instanceof GroundRequest => "ground request",
   default => "default request"
};