PHP - 避免编写嵌套的 if 语句

如何解决PHP - 避免编写嵌套的 if 语句

我有这个变量：

$otcId;

可以从 3 个不同位置检索哪个值（请记住，该值在所有位置始终相同）：

$otcId = $this->dat['id_rfcOV'];
$otcId = $this->request['id_rfcOV'];
$otcId = $this->response['id_rfcOV'];

我的方法如下：

$otcId = $this->dat['id_rfcOV'];

if(isset($this->dat['id_rfcOV'])){
   $otcId = $this->dat['id_rfcOV'];
} elseif(isset($this->request['id_rfcOV'])) {
   $otcId = $this->request['id_rfcOV'];
} elseif(isset($this->response['id_rfcOV'])) {
   $otcId = $this->response['id_rfcOV'];
}

编写这段代码的更短更好、更易读的方法是什么？

解决方法

如果我是你，我会把它包装在一个函数中，并使用 null coalescing operator 来简化检索和默认值的返回。

<?php

class MyController
{
    private array $dat = [];
    private array $request = ['id_rfcOV' => 'foo'];
    private array $response = ['id_rfcOV' => 'bar'];

    /**
     * @param string $name Parameter name
     * @param null $default Default value to return if no matching parameters are found
     * @return mixed|string|null
     */
    function getParam(string $name,$default=null)
    {
        return $this->dat[$name] ?? $this->request[$name] ?? $this->response[$name] ?? $default;
    }
    
    function test(): void
    {
        // Using a member function,we can get our parameter value with a one-liner
        $otcId = $this->getParam('id_rfcOV');

        assert($otcId == 'bar','Value should be from the last array checked');

        printf("Value is %s \n",$otcId);

        $val = $this->getParam('non-existent','wombats');

        assert($val == 'wombats','Value should be the default');

        printf("Value is %s \n",$val);
    }
}

$myController = new MyController();
$myController->test();