拉勒维尔5号的子关系？

Question

目前，我有3款车型，Listing， the & Payment，它们有以下几种关系：

列表

class Listing extends Model {

    public function offers() {
        return $this->hasMany(\App\Models\Offer::class)->orderBy('created_at', 'desc');
    }

}

提供

class Offer extends Model {

    public function payment() {
        return $this->hasOne(\App\Models\Payment::class, 'item_id', 'id')->where('item_type', \App\Models\Offer::class)->where('status', '1');
    }

    public function listing() {
        return $this->belongsTo(\App\Models\Listing::class)->withTrashed();
    }

}

付款

class Payment extends Model {

    public function offer() {
        return $this->belongsTo(\App\Models\Offer::class, 'item_id', 'id')->withTrashed();
    }

}

我如何才能从他们列表模型&直接返回与支付表的关系？

Listing可以有无限的报价，但Listing只能有1最大支付

要找到任何相应的支付信息，我必须查询基于模型中的 listing_id的listing_id，然后访问Offer->payment，当我更希望能够这样做时：

$transaction_id = $id;

$listing = Listing::whereHas('payment', function($q) use ($id) {
        $q->where('transaction_id', $id);
        $q->where('user_id', Auth::user()->id);
})->first();

Answer 1

使用HasManyThrough

public function payments() {
    return $this->hasManyThrough(Payment::class, Offer::class, null, 'item_id')
        ->where('payments.item_type', Offer::class)
        ->where('payments.status', '1')
        ->orderBy('offers.created_at', 'desc');
}