Laravel雄辩关系AVG已经

Question

我已经store表相关product表，有product.rating（INT）1至5

我想找到商店，有平均产量3星级或任何基于我的职务数据

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; 

它总是返回像它忽略我的帖子store-rating值的所有数据

这里是我的模型

<?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

如回答下面我应该使用groupBy和havingRaw使查询工作，但转出，是不是我的唯一的问题代码

我将密码更改为：

$store = Store::with('session'); 
if ((int)$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; 

和

它似乎我不能使用$store = new Store;开始我的模型查询

希望这可以帮助别人。

Answer 1

使用havingRaw

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

不要

忘记组尝试