有沒有辦法在房地產模型之間進行沒有原始部分的查詢？

用戶模型：

<?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();
    }
}

場地型號：

<?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);
    }
}

審查模式：

<?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);
    }
}

所以用戶有很多場地，場地有很多評論。

例如，我想在自己的場地上獲得評論（如果我擁有 id 為 100、101 的場地 - 我想獲得這兩個場地的所有評論）

原始查詢是這樣的：

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

我在用戶模型中在 Laravel 中嘗試過的（不起作用），我也很好奇是否有一種方法，沒有原始部分：

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

uj5u.com熱心網友回復：

如果我正確理解您的模型關系，則HasManyThrough關系應該有效：

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

uj5u.com熱心網友回復：

僅需要原始部分，因為您必須在子查詢的選擇部分中包含外鍵。即使您可能不希望查詢結果中出現 user_id，仍然必須選擇它，Laravel 才能使關系匹配作業。

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

uj5u.com熱心網友回復：

我是這樣做的，但我不太確定這是最好的方法，我愿意接受建議：

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

標籤：php mysql 拉拉维尔 雄辩 laravel-8

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