Question

我有这样的查询：

$users = User::with("purchases")->has("purchases",">",10)->orderBy("id","desc")->get();

我想通过购买计数来订购结果。例如，最高买家应该在顶部，反之亦然。有没有办法做到这一点？

Answer 1

查看this，您可以尝试：

$users = User::with("purchases")
    ->has("purchases",">",10)
    ->orderBy("id","desc")
    ->get()
    ->sortBy(function($user){
        return $user->purchases->count();
    });

不能在sql中对它进行排序，但应该完成这项工作。

Answer 2

如果您想让数据库（而不是PHP）对结果进行排序，则必须使用连接。像这样：

$users = User::with('purchases')
    ->selectRaw('users.*, count(purchases.user_id) as aggregate')
    ->leftJoin('purchases', 'purchases.user_id', '=', 'users.id')
    ->groupBy('purchases.user_id')
    ->having('aggregate', '>', 10)
    ->orderBy('aggregate', 'desc')
    ->get();

它看起来有点乱，因此您可以提取到查询范围并执行以下操作：

$users = User::with('purchases')->orderByHas('purchases', '>', 10, 'DESC')->get();

并将其添加到您的基本模型（或仅用户模型）：

public function scopeOrderByHas($query, $relation, $operator = '>=', $count = 1, $dir = 'DESC') {
    $table = $this->getTable();
    $relation = $this->$relation();
    $relatedTable = $relation->getRelated()->getTable();
    $primaryKey = $this->primaryKey;
    $foreignKey = $relation->getForeignKey();

    return $query->selectRaw("{$table}.*, count({$foreignKey}) as aggregate")
        ->leftJoin($relatedTable, $foreignKey, '=', "{$table}.{$primaryKey}")
        ->groupBy($foreignKey)
        ->having('aggregate', $operator, $count)
        ->orderBy('aggregate', $dir);
}

根据＆＃34;有计数＆＃34;有没有办法订购？一个急切的加载雄辩的查询？

2 个答案: