Laravel 7 - 在数据透视表上查询条件

Question

我正在制作一个 AirBnb 副本，我正在尝试创建一个仅返回赞助房屋的 api。

数据库表

我写的查询找不到任何房子，你能帮我找出错误吗？ 谢谢！

$houses = House::with(['position', 'user', 'messages', 'services', 'visualizations', 'sponsorships'])
         ->whereHas('sponsorships', function($query) {
            $query->whereTime('sponsor_start', '<=', Carbon::now())
                  ->whereTime('sponsor_end', '>', Carbon::now());
         })
         ->where('is_visible','=', 1)
         ->paginate(12);;

Answer 1

sponsor_start和sponsor_end是数据透视表上的列

假设您的关系定义如下：

//Sponsorship model
public function houses()
{
    return $this->belongsToMany(House::class)
        ->withPivot('sponsorship_start', 'sponsorship_end')
        ->withTimestamps();
}

//House model
public function sponsorships()
{
    return $this->belongsToMany(Sponsorship::class)
        ->withPivot('sponsorship_start', 'sponsorship_end')
        ->withTimestamps();
}

您可以尝试以下方法 - 通过数据透视表名称限定数据透视表的列

$houses = House::with(['position', 'user', 'messages', 'services', 'visualizations', 'sponsorships'])
         ->whereHas('sponsorships', function($query) {
            $query->whereTime('house_sponsorship.sponsor_start', '<=', Carbon::now())
                  ->whereTime('house_sponsorship.sponsor_end', '>', Carbon::now());
         })
         ->where('is_visible','=', 1)
         ->paginate(12);;