5

我正在开发一个基于 Laravel 3 构建的项目，我正在尝试查看是否可以缩短处理关系的代码（更好的方法如下）

用户控制器

创建用户函数

$newUser = new User;

if($userData['organization'])
    $newUser->organization = self::_professional('Organization', $newUser, $userData);
else
    $newUser->school = self::_professional('School', $newUser ,$userData);

创建或检索学校/组织 ID

private function _professional($type, $newUser, $userData)
{
    if ( $orgId = $type::where('name', '=', $userData[strtolower($type)])->only('id'))
        return $orgId;
    else
    {
        try {
            $org = $type::create(array('name' => $userData[strtolower($type)]));
            return $org->attributes['id'];
        } catch( Exception $e ) {
            dd($e);
        }
    }
}

楷模

用户模型

class User extends Eloquent {

    public function organization()
    {
        return $this->belongs_to('Organization');
    }

    public function school()
    {
            return $this->belongs_to('School');
    }
}

组织/学校模式

class Organization extends Eloquent {

    public function user() 
    {
        return $this->has_many('User');
    }

}

迁移

用户迁移

....
$table->integer('organization_id')->unsigned()->nullable();
$table->foreign('organization_id')->references('id')->on('organizations');

$table->integer('school_id')->unsigned()->nullable();
$table->foreign('school_id')->references('id')->on('schools');
....

组织/学校迁移

....
$table->increments('id');
$table->string('name');
$table->string('slug');
$table->integer('count')->default(1)->unsigned();
....

现在，我的问题是：

  1. 有没有更好的方法来生成 User -> School/Organization 关系，然后是上面使用的那个？如果是这样，怎么做？

  2. 然后通过执行以下操作来检索用户的学校/组织名称的任何更好的方法：School::find($schoolId)->get()

这样做User::find(1)->school()不会检索任何数据school，仅：

[base:protected] => User Object
(
    [attributes] => Array
        (
            [id] => 1
            [nickname] => w0rldart
            ....
            [organization_id] => 
            [school_id] => 1
            ...
        )
    [relationships] => Array
        (
        )

    [exists] => 1
    [includes] => Array
        (
        )

)

[model] => School Object
(
    [attributes] => Array
        (
        )

    [original] => Array
        (
        )

    [relationships] => Array
        (
        )

    [exists] => 
    [includes] => Array
        (
        )

)
4

2 回答 2

3 
// You have to save this before you can tied the organizations to it
$new_user->save();

// The organizations that you want to tie to your user
$oganization_ids = array(1, 2, 3);

// Save the organizations
$result = $new_user->organization()->sync($oganization_ids);
于 2013-06-10T19:52:57.530 回答
3

有没有更好的方法来生成User -> School/Organization关系，然后是上面使用的那个？如果是这样，怎么做？

Eloquent 应该能够自动处理它，请参见此处：http ://three.laravel.com/docs/database/eloquent#inserting-related-models 。

然后通过执行以下操作来检索用户的学校/组织名称的任何更好的方法：School::find($schoolId)->get()

如果您的意思是您无法检索关系信息：has_name并belongs_to返回一个 Eloquent 表达式，您仍然需要调用get()它们来检索您想要的对象或集合。

我处理这个问题的方法是为关系添加一个参数以确定该方法是否应该返回表达式或结果：

$table->increments('id');
$table->string('name');
$table->string('slug');
$table->integer('count')->default(1)->unsigned();
$table->unique('name');

...

class User extends Eloquent
{
    /**
     * Return an Elloquent Expression or a collection (or object)
     * Defaults to return a collection
     */
    public function organization ($as_expression=FALSE) {
        $related = $this->belongs_to('Organization');
        return $as_expression ? $related : $related->get();
    }

    public function school ($as_expression=FALSE) {
        $related = $this->belongs_to('School');
        return $as_expression ? $related : $related->get();
    }

    ...

    public function create ($user_data=array()) {
        $newUser = new User;
        $activity = @$user_data['organization'] ?
            'organization' : (@$user_data['school'] ? 'school' : '');

        switch ($activity) {
            case 'organization':
            case 'school':
                /**
                 * Call the corresponding method, return as Eloquent Expression
                 * Use a string to be verbose, TRUE does fine if wanted
                 */
                $new_user->{$activity}('as_expression')->save(array( 'name' => $user_data[$activity]));

                break;
            default:
                break;
        }

        return $new_user;
    }
}

...

class Organization extends Eloquent
{
    public function user ($as_expression=FALSE) {
        $related = $this->has_many('User');
        return $as_expression ? $related : $related->get();
    }
}
于 2013-07-01T08:06:48.553 回答