如何在 laravel 查询构建器中为同一列构建多个条件的查询？

Question

我想从 'user' table 获取计数，其中 'plan_validity' 为空或大于今天的日期。这两个条件都需要检查 user_id 和 status = 1.

的同一列

  $currentUser = User::where('user_id', $card_details->user_id)
                ->where('status', 1)
                ->where(function (Builder $query) {
                    return $query
                        ->whereDate('plan_validity', '>=', Carbon::now())
                        ->orWhere('plan_validity', null);
                    })->count();

给出错误

[2021-11-23 10:40:31] production.ERROR: Argument 1 passed to App\Http\Controllers\ProfileController::App\Http\Controllers\{closure}() must be an instance of App\Http\Controllers\Builder, instance of Illuminate\Database\Eloquent\Builder given, called in /home/hellovcard/public_html/vendor/laravel/framework/src/Illuminate/Database/Eloquent/Builder.php on line 237 {"userId":17,"exception":"[object] (TypeError(code: 0): Argument 1 passed to App\Http\Controllers\ProfileController::App\Http\Controllers\{closure}() must be an instance of App\Http\Controllers\Builder, instance of Illuminate\Database\Eloquent\Builder given, called in /home/hellovcard/public_html/vendor/laravel/framework/src/Illuminate/Database/Eloquent/Builder.php on line 237 at /home/hellovcard/public_html/app/Http/Controllers/ProfileController.php:34)

如何修改上述查询？

这是我需要得到的输出

enter image description here

Answer 1

您没有正确组合“where”子句。在末尾使用“orWhere”，它基本上忽略了前面 where 子句的所有条件（即它与哪个用户相关），因此将 plan_validity 为空的任何行添加到结果中。因此，在按用户 ID 和状态查询后，您需要将查询的其余部分组合成一个逻辑组，如下所示：

$user = Users::where('user_id', $card_details->user_id)
  ->where('status', 1)
  ->where(function (Builder $query) {
        return $query
            ->whereDate('plan_validity', '>=', Carbon::now())
            ->orWhere('plan_validity', null);
    })
    ->get();

如何在 laravel 查询构建器中为同一列构建多个条件的查询？

How to build query for multiple condition for same column in laravel query builder?

mysql

sql

sql-server

laravel-8

laravel-query-builder