将接口发送到构造函数 PHP

Question

我已经下载了一个支付连接示例。不，我正在尝试使用它，但是构造函数想要在我声明 ClassName 时获取接口但我不知道该怎么做。我试过了

$interface = CallbackInterface::class;
$interface = CallbackInterface();
$interface = CallbackInterface;

还有很多，但我想不通。我唯一知道的是用 class 实现一个接口。也许是菜鸟问题，但我搜索了将近一天都没有成功。

$config = new Config('string1', 'string2');
$pay = new ClassName($config, $interface);

interface CallbackInterface
{   
    public function Key($sIdentifier, $sTransactionKey);
    public function tSuccess($sTransactionKey);
}

class ClassName
{
    public function __construct(Config $oConfig, CallbackInterface $oCallbacks)
    {
        $this->oConfig = $oConfig;
        $this->oCallbacks = $oCallbacks;
    }
}

Answer 1

您应该按照这些思路寻找解决方案

// Create a class that implements the interface (e.g. MyClass)
// MyClass implements the interface functions: Key and tSuccess
// MyClass can now be injected as type CallbackInterface into the __construct() of class ClassName

Class MyClass implements CallbackInterface
{
    public function Key($sIdentifier, $sTransactionKey)
    {
        // your implementation here
    }
    public function tSuccess($sTransactionKey)
    {
        // your implementation here
    }
}

interface CallbackInterface
{   
    public function Key($sIdentifier, $sTransactionKey);
    public function tSuccess($sTransactionKey);
}

class ClassName
{
    public function __construct(Config $oConfig, CallbackInterface $oCallbacks)
    {
        $this->oConfig = $oConfig;
        $this->oCallbacks = $oCallbacks;
    }
}

$config = new Config('string1', 'string2');
$interface = new MyClass(); // you've now instantiated an object of type CallbackInterface
$pay = new ClassName($config, $interface);

将接口发送到构造函数 PHP

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