Laravel Eloquent：访问属性和动态表名

I am using the Laravel Framework and this question is directly related to using Eloquent within Laravel.

我试图建立一个Eloquent 的模型，可以在多个不同的表格中使用.原因是我有多个基本相同但每年都不同的表，但我不想重复代码来访问这些不同的表.

• gamedata_2015_nations
• gamedata_2015_leagues
• gamedata_2015_团队
• gamedata_2015_players

我当然可以有一个带有年份栏的大桌子，但由于每年有超过35万行，而且要处理很多年，我决定最好将它们拆分为多个表格，而不是在每个请求上都有额外"where"的4个大桌子.

So what I want to do is have one class for each and do something like this within a Repository class:

public static function getTeam($year, $team_id)
    {
        $team = new Team;

        $team->setYear($year);

        return $team->find($team_id);
    }

I have used this discussion on the Laravel forums to get me started: http://laravel.io/forum/08-01-2014-defining-models-in-runtime

So far I have this:

class Team extends \Illuminate\Database\Eloquent\Model {

    protected static $year;

    public function setYear($year)
    {
        static::$year= $year;
    }

    public function getTable()
    {
        if(static::$year)
        {
            //Taken from https://github.com/laravel/framework/blob/4.2/src/Illuminate/Database/Eloquent/Model.php#L1875
            $tableName = str_replace('\\', '', snake_case(str_plural(class_basename($this))));

            return 'gamedata_'.static::$year.'_'.$tableName;
        }

        return Parent::getTable();
    }
}

这似乎有效，但我担心它没有以正确的方式工作.

因为我使用的是静电关键字，所以属性$Year保留在类中，而不是每个单独的对象中，所以每当我创建新对象时，它仍然持有基于上次在不同对象中设置的$Year属性.我宁愿$Year与单个对象相关联，并且需要在每次创建对象时进行设置.

Now I am trying to track the way that Laravel creates Eloquent models but really struggling to find the right place to do this.

For instance if I change it to this:

class Team extends \Illuminate\Database\Eloquent\Model {

    public $year;

    public function setYear($year)
    {
        $this->year = $year;
    }

    public function getTable()
    {
        if($this->year)
        {
            //Taken from https://github.com/laravel/framework/blob/4.2/src/Illuminate/Database/Eloquent/Model.php#L1875
            $tableName = str_replace('\\', '', snake_case(str_plural(class_basename($this))));

            return 'gamedata_'.$this->year.'_'.$tableName;
        }

        return Parent::getTable();
    }
}

当试图组建单个团队时，这一点工作得很好.然而，在人际关系中，这是行不通的.这是我在人际关系中try 过的:

public function players()
{
    $playerModel = DataRepository::getPlayerModel(static::$year);

    return $this->hasMany($playerModel);
}

//This is in the DataRepository class
public static function getPlayerModel($year)
{
    $model = new Player;

    $model->setYear($year);

    return $model;
}

Again this works absolutely fine if i'm using static::$year, but if I try and change it to use $this->year then this stops working.

The actual error stems from the fact that $this->year is not set within getTable() so that the parent getTable() method is called and the wrong table name returned.

我的下一步是try 找出为什么它使用静电属性，而不是非静态属性(不确定用于该属性的正确术语).当我试图建立球员关系时，我假设它只是使用团队类中的静电::$Year.然而，情况并非如此.如果我try 使用类似于以下内容的命令强制执行错误，请执行以下操作:

public function players()
{
    //Note the hard coded 1800
    //If it was simply using the old static::$year property then I would expect this still to work
    $playerModel = DataRepository::getPlayerModel(1800);

    return $this->hasMany($playerModel);
}

现在发生的是，我得到一个错误，说找不到gamedata_1800_玩家.也许这并不奇怪.但它排除了Eloquent只是使用Team类中的static::$year属性的可能性，因为它明确地设置了我发送给getPlayerModel()方法的自定义年份.

所以现在我知道，当$Year在关系中设置并且是静态设置的时候，gettable()可以访问它，但是如果它是非静态设置的，那么它就会丢失在某个地方，并且对象在调用gettable()时不知道这个属性.

(note the significance of it working different when simply creating a new object and when using relationships)

我意识到我已经给出了很多细节，所以为了简化和澄清我的问题:

1) Why does static::$year work but $this->year not work for relationships, when both work when simply creating a new object.

2) Is there a way that I can use a non static property and achieve what I am already achieving using a static property?

理由:静态属性将保留在类中，即使我已经完成了一个对象，并试图用该类创建另一个对象，这似乎是不对的.

Example:

    //Get a League from the 2015 database
    $leagueQuery = new League;

    $leagueQuery->setYear(2015);

    $league = $leagueQuery->find(11);

    //Get another league
    //EEK! I still think i'm from 2015, even though nobodies told me that!
    $league2 = League::find(12);

This may not be the worst thing in the world, and like I said, it is actually working using the static properties with no critical errors. However it is dangerous for the above code sample to work in that way, so I would like to do it properly and avoid such a danger.