Php 如何将新变量与类中的默认变量相匹配?

Php 如何将新变量与类中的默认变量相匹配?,php,settings,extend,Php,Settings,Extend,在jQuery中有$.extend()函数。这基本上类似于将默认设置与输入设置相匹配。我想在PHP中使用类似的技术,但似乎没有类似的函数。所以我想知道:是否有类似于.extend()的函数,但在PHP中?如果没有,那么,还有什么替代方法? 这是一个示例类,以及我当前如何获得这种效果。我还添加了一条评论,我希望如何做到这一点: class TestClass { // These are the default settings: var $settings = array(

在jQuery中有$.extend()函数。这基本上类似于将默认设置与输入设置相匹配。我想在PHP中使用类似的技术,但似乎没有类似的函数。所以我想知道:是否有类似于
.extend()
的函数,但在PHP中?
如果没有,那么,还有什么替代方法?

这是一个示例类,以及我当前如何获得这种效果。我还添加了一条评论,我希望如何做到这一点:

class TestClass {
    // These are the default settings:
    var $settings = array(
        'show' => 10,
        'phrase' => 'Miley rocks!',
        'by' => 'Kalle H. Väravas',
        'version' => '1.0'
    );
    function __construct ($input_settings) {
        // This is how I would wish to do this:
        // $this->settings = extend($this->settings, $input_settings);

        // This is what I want to get rid of:
        $this->settings['show'] = empty($input_settings['show']) ? $this->settings['show'] : $input_settings['show'];
        $this->settings['phrase'] = empty($input_settings['phrase']) ? $this->settings['phrase'] : $input_settings['phrase'];
        $this->settings['by'] = empty($input_settings['by']) ? $this->settings['by'] : $input_settings['by'];
        $this->settings['version'] = empty($input_settings['version']) ? $this->settings['version'] : $input_settings['version'];

        // Obviously I cannot do anything neat or dynamical with $input_settings['totally_new'] :/
    }
    function Display () {
        // For simplifying purposes, lets use Display and not print_r the settings from construct
        return $this->settings;
    }
}

$new_settings = array(
    'show' => 30,
    'by' => 'Your name',
    'totally_new' => 'setting'
);

$TC = new TestClass($new_settings);

echo '<pre>'; print_r($TC->Display()); echo '</pre>';
class测试类{
//以下是默认设置:
var$settings=array(
“显示”=>10,
“短语”=>“麦莉洛克!”,
“作者”=>“Kalle H.Väravas”,
“版本”=>“1.0”
);
函数构造($input\U设置){
//我希望这样做:
//$this->settings=extend($this->settings,$input\u settings);
//这就是我想要摆脱的:
$this->settings['show']=空($input_settings['show'])?$this->settings['show']:$input_settings['show'];
$this->settings['phrase']=空($input_settings['phrase'])?$this->settings['phrase']:$input_settings['phrase'];
$this->settings['by']=空($input_settings['by'])?$this->settings['by']:$input_settings['by'];
$this->settings['version']=空($input_settings['version'])?$this->settings['version']:$input_settings['version'];
//显然,我无法使用$input_设置[“全新”]:/
}
功能显示(){
//为了简化起见,让我们使用“显示”而不是“打印”构造中的设置
返回$this->settings;
}
}
$new\u设置=数组(
“显示”=>30,
'按'=>'您的姓名',
“全新”=>“设置”
);
$TC=新的测试类($new_设置);
回声';打印($TC->Display());回声';

如果您注意到,有一个全新的设置:
$new\u settings['all\u new']
。这应该作为
$this->settings['otherly_new']
包含在数组中。PS:以上代码输出。

尝试使用array\u merge php函数。代码:


我再一次觉得自己很愚蠢。回答得真好<代码>$this->settings=array\u merge($this->settings,$input\u settings)做了超出我预期的把戏:)奇怪的是,就在大约一小时前,我的眼睛滑过了数组_merge函数^^。谢谢你,伙计!
array_merge($this->settings, $input_settings);