从不同关系中获取最新信息雄辩的laravel

Beg*_*ner 2 php laravel eloquent laravel-5

我有3张桌子

模型网址

class Url extends Model
{
    public function users(){
        return $this->belongsToMany(User::class);
    }
    public function url_status(){
        return $this->hasMany(UrlStatus::class);
    }
}
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型号UrlStatus

class UrlStatus extends Model
{
    public function url()
    {
        return $this->belongsTo(Url::class);
    }
}
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模型使用者

class User extends Authenticatable
{
    use Notifiable, SoftDeletes, HasRoles;
    public function urls(){
        return $this->belongsToMany(Url::class);
    }
}
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在我的控制器中,我正在查询:

$url = Url::with('url_status','users')->where('list_status', true)->get();
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如何获取最新的url_status?

编辑-

这是迁移文件中每个表的结构

网址表

 Schema::create('urls', function (Blueprint $table) {
            $table->increments('id');
            $table->string('url');
            $table->string('description');
            $table->boolean('list_status');
            $table->timestamps();
        });
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网址状态

Schema::create('url_status', function (Blueprint $table) {
            $table->increments('id');
            $table->unsignedInteger('url_id');
            $table->integer('status_code');
            $table->string('status');
            $table->boolean('sms');
            $table->boolean('email');
            $table->timestamps();
        });
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用户表

Schema::create('users', function (Blueprint $table) {
            $table->increments('id');
            $table->string('name');
            $table->string('email')->unique();
            $table->string('contacts');
            $table->string('password');
            $table->rememberToken();
            $table->timestamps();
            $table->softDeletes();
        });
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rkj*_*rkj 5

您也可以这样尝试

模型网址

public function latestUrlStatus(){
    return $this->hasOne(UrlStatus::class)->latest();
}
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像这样获取

$urls = Url::with('latestUrlStatus','users')->where('list_status', true)->get();
foreach($urls as $url){
    echo $url->latestUrlStatus->status_code;
}
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