Laravel Eloquent 关系 AVG 有

Question

我有 store table 与 product table 相关，有 product.rating (int) 1 到 5

我想根据我的 post 数据

找到平均产品评级为 3 星或任何的商店

我的代码

$store = new Store;
if ((int)$request->input('store-rating')>0) {
    $store->whereHas('product',function($query) use($request) {
        $rate = $request->input('product-rating');
        $query->where(DB::raw('AVG(rating)'),$rate);
    });
}
$store->with('product');
$thestore = $store->paginate(6);

return $thestore;

它总是 return 所有类似它的数据忽略我的 POST store-rating 值

这是我的模型

<?php

namespace App;

use Illuminate\Database\Eloquent\Builder;
use Illuminate\Database\Eloquent\Model;
use Illuminate\Database\Eloquent\Scope;
use Illuminate\Database\Eloquent\SoftDeletes;
use Illuminate\Foundation\Auth\User as AuthUser;
use Illuminate\Support\Facades\Auth;
use App\ModelBuilder;

class Store extends AuthUser
{
    use SoftDeletes;

    protected $guarded = [];
    protected $dates = ['deleted_at'];

    public function product(){
        return $this->hasMany('App\Product');
    }


}

解决方案

如下所述，我应该使用 groupBy 和 havingRaw 来使查询工作，但事实证明这不是我在代码中唯一的问题

我把代码改成了：

$store = Store::with('session');
if ((int)$request->input('store-rating')>0) {
    $store->whereHas('product',function($query) use($request) {
        $rate = $request->input('product-rating');
        $query->groupBy('rating')->havingRaw('AVG(rating) = '.$rate);
    });
}
$thestore = $store->paginate(6);

return $thestore;

我似乎无法使用 $store = new Store; 来启动我的模型查询

希望这对其他人有帮助。

Answer 1

尝试使用 havingRaw

$query->grouBy('rating')->havingRaw('AVG(rating) = '.$rate);

别忘了分组依据

Laravel Eloquent 关系 AVG 有

Laravel Eloquent Relationship AVG having

php

mysql

laravel

eloquent

laravel-eloquent