Laravel 7 -基于透视表的查询

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。

DB表

我写的查询不能找到任何房子，你能帮我找到错误吗?谢谢你！

$houses = House::with(['position', 'user', 'messages', 'services', 'visualizations', 'sponsorships'])
->whereHas('sponsorships', function($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是透视表中的列

假设你的关系定义如下:

//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();
}

您可以尝试以下操作-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);;