Question

假设有三种等级:

Sponsor id <[int], name [string], 赞助的项目 [字符串], 赞助的项目 [数组 Project 对象]}

project > id [int], name [string], desc [string], succors <[强势>Sponsor 对象的数组]}

" 强 " 学校 < /强'学校 < /强'code > id < /code > [int], name [string], projectsDone [强 " 强 " 项目

我被困在 < strong > sponsor s < code > 赞助的项目和 < strong > Project s < code > 赞助者之间。

从理论上讲,如果我创建一个“强”项目 < / 强”, 那么我也需要为此创建“ 强” 赞助人 < / 强” 。如果我创建了“强” 赞助人 < / 强”, 那么我也需要为此创建“强” 项目 < / 强” 。

因此,这会造成无穷无尽的 " 强 " 项目

如果是这样的话,我该如何避免这一缺陷?

我的目标是从返回的多个 SQL 结果中临时创建所有这些对象, 然后再将它发送到 JavaScript 页面。

final class Sponsor{
    private $_id;
    private $_name;
    private $_projects;

    public function __construct($id, $name, Project $projects){
        $this->setSponsor($id, $name, $projects);
    }

    public function getId(){
        return $this->_id;
    }

    public function getName(){
        return $this->_name;
    }

    public function getProjects(){
        return $this->_projects;
    }

    private function setId($id){
        $this->_id = $id;
    }

    private function setName($name){
        $this->_name = $name;
    }

    private function setProject(Project $projects){
        $this->_projects = array();
        foreach($projects as $val){
            array_push($this->_projects, $val);
        }
    }

    private function setSponsor($id, $name, Project $projects){
        $this->setId($id);
        $this->setName($name);
        $this->setProjects($projects);
    }
}

final class Project{
    private $_id;
    private $_name;
    private $_desc;
    private $_sponsors;

    public function __construct($id, $name, $desc, Sponsor $sponsors){
        $this->setProject($id, $name, $desc, $sponsors);
    }

    private function setId($id){
        $this->_id = $id;
    }

    private function setName($name){
        $this->_name = $name;
    }

    private function setDesc($desc){
        $this->_desc = $desc;
    }

    private function setSponsors(Sponsor $sponsors){
        $this->_sponsors = array();
        foreach($sponsors as $obj){
            $id = $obj->getId();
            $name = $obj->getName();
            $projects = $obj->getProjects();
            array_push($this->_sponsors, new Sponsor($id, $name, Project $this));
        }
    }

    private function setProject($id, $name, $desc, Sponsor $sponsors){
        $this->setId($id);
        $this->setName($name);
        $this->setDesc($desc);
        $this->setSponsors($sponsors);
    }
}

final class School{

    private $_id;
    private $_name; 
    private $_projectsDone;

    public function __construct($id, $name, Project $projects){
        $this->setSchool($id, $name, $projects);
    }

    private function setId($id){
        $this->_id = $id;
    }

    private function setName($name){
        $this->_name = $name;
    }

    private function setProjectsDone($projects){
        $this->_projectsDone = array();
        foreach($projectsDone as $val){
            $this->{_projectsDone}[] = $val;
        }
    }

    public function setSchool($id, $name, $projectsDone){
        $this->setId($id);
        $this->setName($name);
        $this->setProjectsDone($projectsDone);
    }
}

Answer 1

您已经有了一个 < a href= > 的问题。 “ http:// en.wikipedia. org/ wiki/ circulal_ reference" rel= “ nofollow” > circural reference 。每一个对象都持有与其相关的物体阵列, 尝试保留其代号的阵列, 并在需要时查找相关实体。

Answer 2

物体相互之间互相保持联系是没有错的。它非常常见, 并且是一种正常( 弹性) 的关系。比如, 标注中的 DOM. 元素节点知道自己的孩子, 其孩子知道自己的父母。

显然,在建造和初始化它们时,你需要小心小心, 因为有一点点鸡和鸡amp; 鸡蛋问题, 但支票归结到

if relation to object not established
    establish()

您提到过您将此信息存储在数据库中。您可能甚至可以避免此检查。例如, 您可能拥有以下表格 :

Sponsor
Project
School
SponsoredProjects - many to many relation
SchoolProjectsDone - many to many relation

构造赞助者、项目和学校对象,然后在赞助项目和学校项目表格中环绕关系,将对象联系起来。

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