拉拉维尔渴望从有关系的负载模型

Question

有3个表：

users： ID， 姓名

compliments： ID， 姓名

users_compliments： id， compliment_id， user_id， complimentor_id

这是我的用户模型：

class User extends Model
public function sent_compliments()  {  
return 
$this->belongsToMany(
'Compliment', 
'users_compliments', 
'complimentor_id',
'compliment_id'
)->withPivot('user_id');
}

如果我运行$user->sent_compliments()->get()，这是一个示例响应

IlluminateDatabaseEloquentCollection {
all: [
Compliment {
id: 2,
name: "You are smart",
pivot: IlluminateDatabaseEloquentRelationsPivot {
user_id: 6,
id: 99,
},
},
Compliment {
id: 52,
name: "You are funny",
pivot: IlluminateDatabaseEloquentRelationsPivot {
user_id: 42,
id: 186,
},
},
],
}

但这是我想要的响应(它包括实际的用户模型，而不仅仅是枢轴中的user_id(

IlluminateDatabaseEloquentCollection {
all: [
Compliment {
id: 2,
name: "You are smart",
pivot: IlluminateDatabaseEloquentRelationsPivot {
user_id: 6,
id: 99,
},
user {
id: 6,
name: "John"
}
},
Compliment {
id: 52,
name: "You are funny",
pivot: IlluminateDatabaseEloquentRelationsPivot {
user_id: 42,
id: 186,
},
user {
id: 42,
name: "Alex"
}
},
],
}

有没有办法通过雄辩的急切加载(例如使用With::命令(获取此数据，而不是遍历透视、收集user_ids并手动查询User表？

Answer 1

您需要定义pivot类并将关系belongsTo添加到透视模型中user_id

用户模型

public function sent_compliments()  {  
return 
$this->belongsToMany(
'Compliment', 
'users_compliments', 
'complimentor_id',
'compliment_id'
)->withPivot('user_id')->using('AppUserComplement');
}

枢轴模型使用 Illuminate\Database\Eloquent\Relations\Pivot;

class UserComplement extends Pivot {
protected $table = 'users_compliments';
public function user(){
return $this->belongsTo('AppUser', 'user_id');
}
}

费斯数据

//if you have user instance already then
$complements = $user->sent_compliments;
//if you don't have fetched user yet then try this 
$user = User::with('sent_compliments')->find($userId);
$complements = $user->sent_compliments;
foreach($complements as $complement){
dd($complement);
dd($complement->pivot->user);
}

有关详细信息，请查看此定义自定义中间表模型https://laravel.com/docs/5.6/eloquent-relationships#many-to-many

编辑

正如@JonasStaudenmeir指出的那样，没有办法急于加载pivot user关系。这是另一个解决方案

用户模型

public function sent_compliments()  {  
return $this->hasMany(UserCompliment::class , 'complimentor_id');
}

用户补充模型

class UserComplement extends Model {
$table = 'users_compliments';
public function complimentor()  {  
return $this->belongsTo(User::class , 'complimentor_id');
}
public function complement()  {  
return $this->belongsTo(Compliment::class , 'compliment_id');
}
public function user()  {  
return $this->belongsTo(User::class , 'user_id');
}
}

获取数据

$user = User::with('sent_compliments','sent_compliments.complement', 'sent_compliments.user')->find($userId);
$sentComplements = $user->sent_compliments;
foreach($sentComplements as $sentComplement){
dd($sentComplement->complement);
dd($sentComplement->user);
}

希望它能清除你