有没有办法在 realtion 模型之间进行没有原始部分的查询？

Question

用户模型：

<?php

namespace App\Models;

use Illuminate\Contracts\Auth\MustVerifyEmail;
use Illuminate\Database\Eloquent\Collection;
use Illuminate\Database\Eloquent\Factories\HasFactory;
use Illuminate\Foundation\Auth\User as Authenticatable;
use Illuminate\Notifications\Notifiable;
use Laravel\Sanctum\HasApiTokens;

class User extends Authenticatable
{
    use HasApiTokens, HasFactory, Notifiable;

    /**
     * The attributes that are mass assignable.
     *
     * @var string[]
     */
    protected $fillable = [
        'name',
        'email',
        'password',
    ];

    /**
     * The attributes that should be hidden for serialization.
     *
     * @var array
     */
    protected $hidden = [
        'password',
        'remember_token',
    ];

    /**
     * The attributes that should be cast.
     *
     * @var array
     */
    protected $casts = [
        'email_verified_at' => 'datetime',
    ];

    public function venues()
    {
        return $this->hasMany(Venue::class);
    }

    public function reviews()
    {
        return $this->hasMany(Review::class);
    }

    public function profile()
    {
        return $this->hasOne(Profile::class);
    }

    public function approvedVenues()
    {
        return $this->hasMany(Venue::class)->where('is_approved', '=', 1);
    }

    public function unapprovedVenues()
    {
        return $this->hasMany(Venue::class)->where('is_approved', false);
    }

    public function ownVenuesReviews()
    {
        return $this->reviews()->whereIn('user_id', function($query) {
            $query->select('id')
            ->from('venues')
            ->whereRaw('venues.user_id = users.id');
        })->get();
    }
}

场地模型：

<?php

namespace App\Models;

use Illuminate\Database\Eloquent\Factories\HasFactory;
use Illuminate\Database\Eloquent\Model;

class Venue extends Model
{
    use HasFactory;

    protected $fillable = [
        'user_id',
        'city_id',
        'category_id',
        'title',
        'address',
        'phone',
        'email',
        'website',
        'facebook',
        'instagram',
        'content_bg',
        'content_en',
        'cover_image',
        'lat',
        'lng'
    ];

    public function user()
    {
        return $this->belongsTo(User::class, 'user_id');
    }

    public function category()
    {
        return $this->belongsTo(Category::class, 'category_id');
    }

    public function city()
    {
        return $this->belongsTo(City::class, 'city_id');
    }

    public function features()
    {
        return $this->belongsToMany(Feature::class, 'venue_feature');
    }

    public function images()
    {
        return $this->hasMany(VenueImage::class);
    }

    public function reviews()
    {
        return $this->hasMany(Review::class);
    }
}

评论模型：

<?php

namespace App\Models;

use Illuminate\Database\Eloquent\Factories\HasFactory;
use Illuminate\Database\Eloquent\Model;

class Review extends Model
{
    use HasFactory;

    protected $fillable = ['rating', 'content', 'venue_id', 'user_id'];

    public function venue()
    {
        return $this->belongsTo(Venue::class);
    }

    public function images()
    {
        return $this->hasMany(ReviewImage::class);
    }

    public function user()
    {
        return $this->belongsTo(User::class);
    }
}

所以用户有很多场地，场地有很多评论。

例如，我想获得对自己场地的评论（如果我拥有 ID 为 100、101 的场地 - 我想获得这两个场地的所有评论）

原始查询是这样的：

SELECT * FROM `reviews` WHERE reviews.venue_id IN (SELECT venues.id FROM venues WHERE venues.user_id = 1)

我在 Laravel 的用户模型中尝试过的方法（不起作用），我也很好奇是否有办法，没有原始部分：

public function ownVenuesReviews()
{
    return $this->reviews()->whereIn('user_id', function($query) {
        $query->select('id')
        ->from('venues')
        ->whereRaw('venues.user_id = users.id');
    })->get();
}

Answer 1

只需要原始部分，因为您必须在子查询的 select 部分中包含外键。即使您可能不希望查询结果中出现 user_id，它仍然必须被 select 编辑以便 Laravel 能够使关系匹配工作。

public function ownVenuesReviews()
{
    return $this->reviews()->whereIn('user_id', function($query) {
        $query->select('id', 'user_id')
        ->from('venues');
    })->get();
}

Answer 2

我是这样做的，但我不太确定，这是最好的方法，我愿意接受建议：

public function ownVenuesReviews()
{
    return Review::whereIn('venue_id', function($query) {
        $query->select('id')
        ->from('venues')
        ->where('user_id', $this->id);
    })->get();
}

Answer 3

如果我正确理解了您的模型关系，HasManyThrough relationship 应该可以工作：

public function ownVenueReviews(): HasManyThrough
{
    return $this->hasManyThrough(Review::class, Venue::class);
}

有没有办法在 realtion 模型之间进行没有原始部分的查询？

Is there a way to make this query without raw part, between realtion models?

php

mysql

laravel

eloquent

laravel-8