Php 类被实例化为对象后，我可以传递_构造参数吗？

我有一个类似的代码片段，如下所示

class Search
{
    public function search($for, $regEx, $flag) //I would like this to be the constructor
    {
        // logic here
        return $this;
    }
}

$this->search($params);
// I have my methods chained, so I could use it in one line like
// $this->search($params)->hasResults();

if ($this->search->hasResults()) {
    echo 'found stuff';
} else {
    echo 'didn't find anything';
}

$this->search->search($params);

if ($this->search->hasResults()) {
    echo 'found stuff';
} else {
    echo 'didn't find anything';
}

public function __call($name, $params)
{
    $call = ucfirst($name);
    $this->$name = new $call($params);
}

然后我有另一个类，它从中创建了一个对象，然后尝试使用该对象

class MyClass
{
    public function start()
    {
        $this->search = new Search();
    }

    public function load()
    {
        $this->search($for, $regEx, $flag);
    }
}

我的问题是，是否可以先创建一个对象，然后给它参数

我知道有一些方法可以解决这个问题，但我问这个问题只是因为我想使用这样的对象

class Search
{
    public function search($for, $regEx, $flag) //I would like this to be the constructor
    {
        // logic here
        return $this;
    }
}

$this->search($params);
// I have my methods chained, so I could use it in one line like
// $this->search($params)->hasResults();

if ($this->search->hasResults()) {
    echo 'found stuff';
} else {
    echo 'didn't find anything';
}

$this->search->search($params);

if ($this->search->hasResults()) {
    echo 'found stuff';
} else {
    echo 'didn't find anything';
}

public function __call($name, $params)
{
    $call = ucfirst($name);
    $this->$name = new $call($params);
}

按照我现在设置的方式，我需要像这样使用它

class Search
{
    public function search($for, $regEx, $flag) //I would like this to be the constructor
    {
        // logic here
        return $this;
    }
}

$this->search($params);
// I have my methods chained, so I could use it in one line like
// $this->search($params)->hasResults();

if ($this->search->hasResults()) {
    echo 'found stuff';
} else {
    echo 'didn't find anything';
}

$this->search->search($params);

if ($this->search->hasResults()) {
    echo 'found stuff';
} else {
    echo 'didn't find anything';
}

public function __call($name, $params)
{
    $call = ucfirst($name);
    $this->$name = new $call($params);
}

我有一个名为search的方法，它执行逻辑，我不想在命名中重复，也不想更改方法的名称

我知道另一种保持视觉吸引力的方法我可以这样传递一个变量

$search=$this->search->search$params

然后

$search->hassresults

同时，我试图向自己介绍新的OOP概念，并从中学习。这需要通过引用传递东西吗？或者设置某种魔法方法？

是的，你可以

class Example {
  public $any;
  function __counstruct($parameters,$some_text) {
     $this->any=$some_text;
     return $this->any;
  }
}

您可以调用构造函数：

$obj = new Example (true,'hello');
echo $obj->any;
$obj->__construct(true,'bye-bye');
echo $obj->any;

是的，这是可能的

请参见下面的示例

<?php
class a{
    public $a = 5;

    public function __construct($var){
        $this->a = $var;
    }
}

$delta = new a(10);
echo $delta->a."\n";
$delta->__construct(15);
echo $delta->a."\n";

---

我能够通过使用uu call magic方法创建我想要的可视化编码，就像这样

class Search
{
    public function search($for, $regEx, $flag) //I would like this to be the constructor
    {
        // logic here
        return $this;
    }
}

$this->search($params);
// I have my methods chained, so I could use it in one line like
// $this->search($params)->hasResults();

if ($this->search->hasResults()) {
    echo 'found stuff';
} else {
    echo 'didn't find anything';
}

$this->search->search($params);

if ($this->search->hasResults()) {
    echo 'found stuff';
} else {
    echo 'didn't find anything';
}

public function __call($name, $params)
{
    $call = ucfirst($name);
    $this->$name = new $call($params);
}

从那里我可以用这个

$this->test->search($params);
$this->test->search->hasResults();

---

当然，我现在将搜索方法设置为类构造函数

，而之前的Anwser显示您可以，我不会使用它，因为它打破了封装的概念。实现你想要的目标的正确方法如下

class Search
{
  public function __constructor($for='', $regEx='', $flag='')
  {
      $this->Setup($for, $regEx, $flag);
  }

   public function Setup($for, $regEx, $flag)
    {
        //assign params
        //clear last result search

        //chain
        return $this;
    }

 public function search()
 {
    // logic here
    return $this;
 }
}

---

通过这种方式，您可以重用对象并在构造函数中使用参数，而不会破坏封装。

执行上述代码并进行检查。它在5.3.x版本的php中工作