我有以下模式
.onCommand()
如何在laravel中正确设置模型?对于attribute_record表,属性和记录之间的关系中存在一个值。我想知道是否需要为attribute_record表建立模型。
我希望能够做一些可以让$ record获取属性及其值的事情。
CREATE TABLE `attributes` (
`id` bigint(20) unsigned NOT NULL AUTO_INCREMENT,
`name` varchar(191) COLLATE utf8mb4_unicode_ci NOT NULL,
`created_at` timestamp NULL DEFAULT NULL,
`updated_at` timestamp NULL DEFAULT NULL,
`deleted_at` timestamp NULL DEFAULT NULL,
PRIMARY KEY (`id`)
) ENGINE=InnoDB AUTO_INCREMENT=5 DEFAULT CHARSET=utf8mb4 COLLATE=utf8mb4_unicode_ci;
CREATE TABLE `records` (
`id` bigint(20) unsigned NOT NULL AUTO_INCREMENT,
`event_school_id` bigint(20) unsigned NOT NULL,
`athlete_id` bigint(20) unsigned NOT NULL,
`year` int(10) unsigned NOT NULL,
`place` int(10) unsigned NOT NULL,
`created_at` timestamp NULL DEFAULT NULL,
`updated_at` timestamp NULL DEFAULT NULL,
`deleted_at` timestamp NULL DEFAULT NULL,
PRIMARY KEY (`id`),
KEY `records_event_school_id_foreign` (`event_school_id`),
KEY `records_athlete_id_foreign` (`athlete_id`),
CONSTRAINT `records_athlete_id_foreign` FOREIGN KEY (`athlete_id`) REFERENCES `athletes` (`id`),
CONSTRAINT `records_event_school_id_foreign` FOREIGN KEY (`event_school_id`) REFERENCES `event_school` (`id`)
) ENGINE=InnoDB AUTO_INCREMENT=4 DEFAULT CHARSET=utf8mb4 COLLATE=utf8mb4_unicode_ci;
CREATE TABLE `attribute_record` (
`id` bigint(20) unsigned NOT NULL AUTO_INCREMENT,
`attribute_id` bigint(20) unsigned NOT NULL,
`record_id` bigint(20) unsigned NOT NULL,
`value` decimal(8,2) NOT NULL,
`created_at` timestamp NULL DEFAULT NULL,
`updated_at` timestamp NULL DEFAULT NULL,
`deleted_at` timestamp NULL DEFAULT NULL,
PRIMARY KEY (`id`),
KEY `attribute_record_attribute_id_foreign` (`attribute_id`),
KEY `attribute_record_record_id_foreign` (`record_id`),
CONSTRAINT `attribute_record_attribute_id_foreign` FOREIGN KEY (`attribute_id`) REFERENCES `attributes` (`id`),
CONSTRAINT `attribute_record_record_id_foreign` FOREIGN KEY (`record_id`) REFERENCES `records` (`id`)
) ENGINE=InnoDB AUTO_INCREMENT=7 DEFAULT CHARSET=utf8mb4 COLLATE=utf8mb4_unicode_ci;
这是我到目前为止所拥有的。
foreach($record->attributes as $attr)
{
echo $attr->value;
}
答案 0 :(得分:1)
我想知道是否需要为attribute_record表建立模型。
不,这是数据透视表,当您正确设置关系时,Laravel将透明地使用它。我假设这是一个many-to-many relationship(记录可以具有许多属性,并且许多记录可以具有相同属性),所以定义您的关系,其余的由Laravel完成:
<?php
namespace App;
use Illuminate\Database\Eloquent\Model;
use Illuminate\Database\Eloquent\SoftDeletes;
class Attribute extends Model
{
use SoftDeletes;
protected $guarded = ['id'];
public function records()
{
return $this->belongsToMany('\\App\\Record')
->withPivot('value')
->withTimestamps();
}
}
<?php
namespace App;
use Illuminate\Database\Eloquent\Model;
use Illuminate\Database\Eloquent\SoftDeletes;
class Record extends Model
{
use SoftDeletes;
protected $guarded = ['id'];
public function attributes()
{
return $this->belongsToMany('\\App\\Attributes')
->withPivot('value')
->withTimestamps();
}
}
现在可以在控制器中执行以下操作:
$record = \App\Record::find($id);
foreach ($record->attributes as $attribute) {
// $attribute is an instance of \App\Attribute
// to access the values in the pivot table, use the pivot attribute
echo $attribute->pivot->value;
}
答案 1 :(得分:1)
<?php
namespace App;
use Illuminate\Database\Eloquent\Model;
use Illuminate\Database\Eloquent\SoftDeletes;
class Record extends Model
{
use SoftDeletes;
protected $guarded = ['id'];
public function attributes()
{
return $this->belongsToMany(Attribute::class, 'attribute_record')
->withPivot('value');
}
}
然后您可以访问以下值:
foreach($record->attributes as $attribute) {
echo $attribute->pivot->value;
}
使用withPivot
方法指定值时,可以在$relatedInstance->pivot->yourColumn
下访问它们。
枢轴值是从中间关系表(在您的示例中为attribute_record
)中检索的