我如何与 Laravel 结合使用多个表？

Question

例如;

 $hall_templates = HallTemplate::where('hall_id' , $id)->orderBy('order' , 'ASC')->get();

输出：

 [{"id":16,"hall_id":3,"template_name":"Magazin Outlet","order":1,"width":12,"height":12},
  {"id":17,"hall_id":3,"template_name":"Merinos K\u00fclt\u00fcr Merkezi","order":2,"width":22,"height":12}]

例如下面的第二个表信息；

$armchair_number_count = HallTemplateArmChair::where('template_id' , 16)->get();

输出：

[{"id":1,"template_id":16,"armchair_no":"a1"}, 
 {"id":2,"template_id":16,"armchair_no":null}]

如何组合这两个表。 我想做输出：

array({"id":16,"hall_id":3,"template_name":"Magazin Outlet","order":1,"width":12,"height":12, "armchair_number_count",2},
{"id":17,"hall_id":3,"template_name":"Merinos K\u00fclt\u00fcr Merkezi","order":2,"width":22,"height":12, "armchair_number_count",0);

感谢您的帮助。

Answer 1

使用withCount() 方法：

HallTemplate::withCount('armchairNumber')->where('hall_id' , $id)->orderBy('order' , 'ASC')->get();

此方法将计算椅子数量并将armchair_number_count 属性添加到每个HallTemplate 对象。

要使其工作，请在 HallTemplate 模型中定义关系：

public function armchairNumber()
{
    return $this->hasMany(HallTemplateArmChair::class, 'template_id');
}