首先看我的结构代码。我需要获得用户汽车的有限优惠。 我有硬编码 3 包。注册后即可免费使用。银和金。如果用户订阅白银,会看到 8 个优惠;如果订阅黄金,会看到 15 个优惠。如果没有订阅,请参阅 2 个优惠(免费套餐)。我的代码:
用户迁移:
$table->increments('id');
$table->string('name');
$table->string('email')->unique();
$table->timestamp('email_verified_at')->nullable();
$table->string('password');
$table->rememberToken();
$table->timestamps();
用户模型:
public function subscriptions_users(){
return $this->hasMany('App\Subscription');
}
public function cars(){
return $this->hasMany('App\Car');
}
public function offers(){
return $this->hasMany('App\Offer');
}
汽车迁移:
$table->bigIncrements('id');
$table->string('car_type');
$table->string('mark');
$table->string('model');
$table->unsignedInteger('user_id');
$table->foreign('user_id')->references('id')->on('users')->onDelete('cascade');
汽车型号:
public function images() {
return $this->hasMany(CarImages::class);
}
public function user() {
return $this->belongsTo('App\User');
}
public function offer() {
return $this->hasMany('App\Offer');
}
提供迁移:
$table->increments('id');
$table->integer('price');
$table->unsignedInteger('user_id');
$table->unsignedInteger('car_id');
报价模型:
public function car() {
return $this->belongsTo('App\Car');
}
public function user() {
return $this->belongsTo('App\User');
}
订阅迁移:
$table->increments('id');
$table->string('subscription');
$table->integer('numberOfOffers');
订阅模式:
public function users(){
return $this->hasMany('App\User');
}
Subscription_user(包含用户和订阅的数据透视表):
$table->increments('id');
$table->integer('user_id')->unsigned()->index();
$table->foreign('user_id')->references('id')->on('users')->onDelete('cascade'); //foreign key relation
$table->integer('subscription_id')->unsigned()->index();
$table->foreign('subscription_id')->references('id')->on('subscriptions')->onDelete('cascade');
订阅用户模型:
public function users(){
return $this->belongsToMany('App\User');
}
public function subscriptions() {
return $this->belongsToMany('App\Subscription');
}
Everythink 工作正常,只是我需要获得限量版汽车的报价。有限版本是字段 $table->integer('numberOfOffers');在订阅表中。我可能有关系问题。我尝试这个但没有成功:
public function getOffers($carID) {
$car = Car::find($carID);
$offers = $car->offer()->limit(THIS IS PROBLEM I DON'T KNOW HOT TO CONNECT USER WITH numberOfOffers)->get();
return response()->json($offers);
}
最佳答案
当你需要对关系运行查询时,你应该使用这样的方法:
$offers = Car::where('id', $carId)->with(['offer' => function($query) use ($limit){
return $query->take($limit);
}])->get();
关于php - 以有限优惠获取汽车 Laravel,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/55004926/