0

假设有三个类：

发起人{ id[int], name[string], sponsoredProjects[array of Project objects]}

项目{ id[int], name[string], desc[string], [赞助商sponsors对象数组]}

School { id[int], name[string], [项目projectsDone对象数组}

我被困在SponsorsponsoredProjects和Projectsponsors之间。

从理论上讲，如果我创建了一个项目，那么我也需要为其创建赞助商。如果我创建赞助商，那么我也需要为它创建项目。

因此，这将导致无穷无尽的Project with Sponsor in their in their with Project in them with Sponsors and etc. 我是对的吗？同样的问题也应该适用于学校，因为他们有一系列项目，其中有一系列赞助商，有一系列项目等等。

如果是这样，那么我应该如何避免这个缺陷？

我的目标是临时从返回的多个 SQL 结果中创建所有这些对象，然后我将其发送到 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);
    }
}
4

2 回答 2

2

你有一个循环引用问题。与其每个对象都保存与其相关的对象数组，不如尝试保留其 id 数组并在需要时查找相关实体。

于 2012-05-25T16:44:00.157 回答
1

让对象在彼此之间来回保持引用并没有错。这很常见，它是集合上的正常（自反）关系。例如，javascript 中的 DOM……一个 Element 节点知道它的子节点，而它的子节点知道它们的父节点。

显然，在构造和初始化它们时需要小心，因为有一点鸡和蛋的问题，但检查归结为

if relation to object not established
    establish()

您提到您将此信息存储在数据库中。你甚至可以避免检查。例如，您可能有以下表格：

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

构造 Sponsor、Project 和 School 对象，然后遍历 SponsoredProjects 和 SchoolProjectsDone 表中的关系，关联对象。

于 2012-05-25T16:45:53.950 回答