PHP 对象到 JSON : How to create class that will have multiple recursive children?

Question

我需要创建一个 PHP 类，该类将包含该类的多个父子关系，以便转换后的 JSON 字符串看起来与此类似。如果它的数组为空，我怎样才能让“ child ”不出现在 JSON 中？

{   name: "Element Parent",
    code: "000"
    children: [
        {
            name: "Element Child 1"
            code: "001"
            children: [
                {
                    name: "Element Child 1A"
                    code: "001A"
                },
                {
                    name: "Element Child 1B"
                    code: "001B"
                    children: [
                        {
                            name: "Element Child 1BA"
                            code: "001BA"
                        }
                    ]
                }
            ]
        }
        ,
        {
            name: "Element Child 2"
            code: "002"
        }
    ]
}

我正在尝试创建一个可以转换为上述 JSON 字符串的 PHP 类。

<?php

class Element implements \JsonSerializable
{
    private $name;
    private $code;
    
    public function __construct($name, $code, )
    {
        $this->name = $name;
        $this->code = $code;
    }
    
    public function jsonSerialize()
    {
        return get_object_vars($this);
    }
    
    public function toJSON(){
        return json_encode($this);
    }
    
    public $children[] = array(); // to contain Element children class 
}

$element = new Element("Element Parent", 000);

$elementChild1 = new Element("Element Child 1", "001");

$elementChild1A = new Element("Element Child 1A", "001A");

$elementChild1B = new Element("Element Child 1B", "001B");
$elementChild1BA = new Element("Element Child 1BA", "001BA");
$elementChild1B->children[] = $elementChild1BA;

$elementChild1->children[] = $elementChild1A;
$elementChild1->children[] = $elementChild1B;

$element->children[] = elementChild1;

$elementChild2 = new Element("Element Child 2", "002");
$element->children[] = elementChild2;

echo $element->toJSON();

?>

非常感谢。

Answer 1

在您实现的jsonSerialize 函数中，您可以更改序列化行为。在那里，您可以检查是否有任何 child ，并在需要时将他们留在外面。在这种情况下，您最终会得到这样的结果:

public function jsonSerialize() {
  $data = [
    "name" => $this->name,
    "code" => $this->code
  ];

  if(!empty($this->children)) {
    $data["children"] = $this->children;
  }

  return $data;
}