问题描述:

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

I am trying to make an Eloquent model that can be used across the multiple different tables. The reason for this is that I have multiple tables that are essentially identical but vary from year to year, but I do not want to duplicate code to access these different tables.

  • gamedata_2015_nations
  • gamedata_2015_leagues
  • gamedata_2015_teams
  • gamedata_2015_players

I could of course have one big table with a year column, but with over 350,000 rows each year and many years to deal with I decided it would be better to split them into multiple tables, rather than 4 huge tables with an extra 'where' on each request.

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();

}

}

This seems to work, however i'm worried it's not working in the right way.

Because i'm using the static keyword the property $year is retained within the class rather than each individual object, so whenever I create a new object it still holds the $year property based on the last time it was set in a different object. I would rather $year was associated with a single object and needed to be set each time I created an object.

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();

}

}

This works just fine when trying to get a single Team. However with relationships it doesn't work. This is what i've tried with relationships:

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.

My next step was to try and figure out why it was working with the static property but not with the nonstatic property (not sure on the right term for that). I assumed that it was simply using the static::$year from the Team class when trying to build the Player relationship. However this is not the case. If I try and force an error with something like this:

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);

}

Now what happens is that I get an error saying gamedata_1800_players isn't found. Not that surprising perhaps. But it rules out the possibility that Eloquent is simply using the static::$year property from the Team class since it is clearly setting the custom year that i'm sending to the getPlayerModel() method.

So now I know that when the $year is set within a relationship and is set statically then getTable() has access to it, but if it is set non-statically then it gets lost somewhere and the object doesn't know about this property by the time getTable() is called.

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

I realise i've given alot of detail now, so to simplify and clarify my question:

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?

Justification for this: The static property will stay with the class even after I have finished with one object and am trying to create another object with that class, which doesn't seem right.

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.

网友答案:

I assume you know how to navigate the Laravel API / codebase since you will need it to fully understand this answer...

Disclaimer: Even though I tested some cases I can't guarantee It always works. If you run into a problem, let me know and I'll try my best to help you.

I see you have multiple cases where you need this kind of dynamic table name, so we will start off by creating a BaseModel so we don't have to repeat ourselves.

class BaseModel extends Eloquent {}

class Team extends BaseModel {}

Nothing exciting so far. Next, we take a look at one of the static functions in Illuminate\Database\Eloquent\Model and write our own static function, let's call it year. (Put this in the BaseModel)

public static function year($year){
    $instance = new static;
    return $instance->newQuery();
}

This function now does nothing but create a new instance of the current model and then initialize the query builder on it. In a similar fashion to the way Laravel does it in the Model class.

The next step will be to create a function that actually sets the table on an instantiated model. Let's call this one setYear. And we'll also add an instance variable to store the year separately from the actual table name.

protected $year = null;

public function setYear($year){
    $this->year = $year;
    if($year != null){
        $this->table = 'gamedata_'.$year.'_'.$this->getTable(); // you could use the logic from your example as well, but getTable looks nicer
    }
}

Now we have to change the year to actually call setYear

public static function year($year){
    $instance = new static;
    $instance->setYear($year);
    return $instance->newQuery();
}

And last but not least, we have to override newInstance(). This method is used my Laravel when using find() for example.

public function newInstance($attributes = array(), $exists = false)
{
    $model = parent::newInstance($attributes, $exists);
    $model->setYear($this->year);
    return $model;
}

That's the basics. Here's how to use it:

$team = Team::year(2015)->find(1);

$newTeam = new Team();
$newTeam->setTable(2015);
$newTeam->property = 'value';
$newTeam->save();

The next step are relationships. And that's were it gets real tricky.

The methods for relations (like: hasMany('Player')) don't support passing in objects. They take a class and then create an instance from it. The simplest solution I could found, is by creating the relationship object manually. (in Team)

public function players(){
    $instance = new Player();
    $instance->setYear($this->year);

    $foreignKey = $instance->getTable.'.'.$this->getForeignKey();
    $localKey = $this->getKeyName();

    return new HasMany($instance->newQuery(), $this, $foreignKey, $localKey);
}

Note: the foreign key will still be called team_id (without the year) I suppose that is what you want.

Unfortunately, you will have to do this for every relationship you define. For other relationship types look at the code in Illuminate\Database\Eloquent\Model. You can basically copy paste it and make a few changes. If you use a lot of relationships on your year-dependent models you could also override the relationship methods in your BaseModel.

View the full BaseModel on Pastebin

网友答案:

Well its not an answer but just my opinion.

I guess, you are trying to scale your application just depending on php part. If you expect that your application will grow by time then it will wise to distribute responsibilities amount all other components. Data related part should handled by RDBMS. As for example if you are using mysql, you can easily partitionize your data by YEAR. And there are lot's of other topic which will help you to manage your data effectively.

网友答案:

Maybe, a custom Constructor is the way to go.

Since all that varies is the year in the name of the corresponding db, your models could implement a constructor like the following:

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

    public function __construct($attributes = [], $year = null) {
        parent::construct($attributes);

        $year = $year ?: date('Y');

        $this->setTable("gamedata_$year_teams");
    }

    // Your other stuff here...

}

Haven't tested this though... Call it like that:

$myTeam = new Team([], 2015);
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