如何在PHP OOP函数中传递对象数组

Question

您好我有一个班级作为孩子，一个班级作为家长。 一个班级再次作为作家信息。

在writer类函数中传递两个对象（obj。来自两个类子）的数组时出错。 我明白了：

Catchable fatal error: Argument 1 passed to PersonWriter::setData() must be an instance of
Person, array given, called in C:\xampp\htdocs\php\oop\book\Person.php on line 62 and
defined in C:\xampp\htdocs\php\oop\book\Person.php on line 40

我的问题是：

1. 我有一个子类，它是父类的直接后代。 我使用Parent类类型创建方法。 为什么不工作？
2. 如何解决？

这是我的代码：

<?php
    // Class Person
    class Person {
        public $name;
        public $gender;
        public $age;

        public function __construct($name, $gender, $age) {
            $this->name = $name;
            $this->gender = $gender;
            $this->age = $age;
        }

        public function getInfo() {
            $info = "Name : $this->name<br/>Gender : $this->gender<br/>Age : $this->age";   
            return $info;
        }
    }

    // Class Mahasiswa
    class Mahasiswa extends Person {
        public $npm;

        public function __construct($npm, $name, $gender, $age) {
            parent::__construct($name, $gender, $age);  
            $this->npm = $npm;
        }

        public function getInfo() {
            $info = "NPM : $this->npm<br/>";
            $info .= parent::getInfo();
            return $info;
        }
    }

    // Class PersonWriter
    class PersonWriter {
        private $persons = array();

        public function setData(Person $persons) {
            $this->persons[] =  $persons;
        }

        public function write() {
            $str = "";
            foreach ($this->persons as $person) {
                $str = "Name : $person->name<br/>";
                $str .= "Gender : $person->gender<br/>";
                $str .= "Age : $person->age<br/>";
            }

            echo $str;
        }
    }

    // Create 2 objects
    $mhs = new Mahasiswa("201143579091","Fandi Akhmad", "L", 21);
    $mhs2 = new Mahasiswa("201143579092","Annisya Ferronica", "P", 19);

    // Add objects to Array 
    $persons = array($mhs, $mhs2);

    $writer = new PersonWriter();
    $writer->setData($persons);
    $writer->write();
?>

答案：

1. 在下面检查为答案的示例代码中。
2. 哦，好吧，我抓住它。 我知道了解我的函数setData（Person $ persons）因为在我的书中没有说出解释。

现在我将person对象添加到数组中，如：

<?php ...
    $writer->setData($mhs);
    $writer->setData($mhs2);
?>

我编辑了我的功能：

public function write() {
            $str = "";
            foreach ($this->persons as $person) {
                $str = "Name : $person->name<br/>";
                $str .= "Gender : $person->gender<br/>";
                $str .= "Age : $person->age<br/>";

                echo "<br/>";
                echo $str;
            }
        }

它现在有效。

Answer 1

这需要一个Person而不是一个数组;

$writer->setData($persons);

这将有效;

$p = new Person("Jake", "male", 29);
$writer->setData($p);

这是因为setData函数需要Person对象;

public function setData(Person $persons) {
    $this->persons[] =  $persons;
}

如果你想让它接受一个数组呢;

public function setData(array $persons) {
    $this->persons[] =  $persons;
}

或者你可以通过删除提示让它接受你想要的任何东西;

public function setData($persons) {
    $this->persons[] =  $persons;
}

编辑

我假设这不是你的代码，因为如果它是非常明显的，为什么它在传递数组时破坏了。 在先用双脚潜水并挣扎之前，请先了解一下OO的基本知识。 该错误使问题显而易见。 在这里解释;

// You are meant to pass in a Person object here, it then appends that
// to an existing array of Persons
public function setData(Person $persons) {
    $this->persons[] =  $persons;
}

// This array of Persons is then iterated over in this function, this is
// where line 48 is, 

public function write() {
    $str = "";
    // Here the foreach goes over the Persons array
    foreach ($this->persons as $person)
    {
        // But the $person objects are being accessed like objects using the
        // -> operator, so if you pass in an array it will fail because you do
        // no access an array using ->
        $str = "Name : $person->name<br/>";
        $str .= "Gender : $person->gender<br/>";
        $str .= "Age : $person->age<br/>";
    }

    echo $str;
}

您可以将这些行更改为以下内容以访问数组;

$str = "Name : " . $person['name'] . "<br/>";
$str .= "Gender : " . $person['gender'] . "<br/>";
$str .= "Age : " . $person['age'] . "<br/>";