Question

您好我正在尝试构建一个动态来自对象的参数。基本上我使用$key作为列名称，使用$value作为列的值。我认为键是准确的列，因为我自己设置它之前在PHP中传递它。这是我的PHP代码。

//model of the request Array [Object] - contains the filter list along with their key
      $filtered_table = Method::leftJoin('users', 'users.id', '=', 'methods.created_by')
      ->leftJoin('roles', 'roles.id', '=', 'users.role_id')
      ->leftJoin('types', 'types.id', '=', 'methods.type_id');

      foreach ($request->filters as $key => $value) {
        $filtered_table->where($key, '=', $value);
      }//dynamic where parameters

      $filtered_table->get([ 'users.username', 'users.id AS users_id', 'methods.*', 'methods.id AS method_id', 'methods.name AS method_name', 'roles.id AS role_id', 'roles.name AS role_name',
      'types.id AS type_id_typetable', 'types.name AS type_name']);

我在这里要做的是做多个->where子句，以便我可以遍历该对象。但我不确定我的代码的语法。

以下是javascript中的类似行为，我分别使用了$ key和$ value。

 for (var i=0; i < filtered_table.length; i++){
            for (var key in filtered_table[i]){
              console.log(key + " -> " + filtered_table[i][key]);
            }
  }

你能帮助我做一些像php那样的行为然后用它来获得动态的->where参数吗？

修改

我遇到了问题，请看我的截图。

 public function filterDataCal(Request $request) {
      //model of the request Array [Object] - contains the filter list along with their key
      $filtered_table = Method::leftJoin('users', 'users.id', '=', 'methods.created_by')
      ->leftJoin('roles', 'roles.id', '=', 'users.role_id')
      ->leftJoin('types', 'types.id', '=', 'methods.type_id')
      ->where(function ($filtered_table) use ($request->filters) {//this is line 122
                   foreach ($request->filters as $key => $value) {
                         $filtered_table->where($key, '=', $value);
                   }
              })
      ->get([ 'users.username', 'users.id AS users_id', 'methods.*', 'methods.id AS method_id', 'methods.name AS method_name', 'roles.id AS role_id', 'roles.name AS role_name',
      'types.id AS type_id_typetable', 'types.name AS type_name']);
      if($filtered_table) {
        return $filtered_table;//data model when returned Array [Object, Object, .....]
      } else {
        return "";
      }
    }

顺便说一下，结果是否仍然像我在那里所做的那样返回？

Answer 1

事实证明我可以直接在where中使用我的对象数组作为搜索地图，请参阅此处。

  $filtered_table = Method::leftJoin('users', 'users.id', '=', 'methods.created_by')
      ->leftJoin('roles', 'roles.id', '=', 'users.role_id')
      ->leftJoin('types', 'types.id', '=', 'methods.type_id')
      ->where($request->filters)    
      ->get([ 'users.username', 'users.id AS users_id', 'methods.*', 'methods.id AS method_id', 'methods.name AS method_name', 'roles.id AS role_id', 'roles.name AS role_name',
      'types.id AS type_id_typetable', 'types.name AS type_name']);

参考

Laravel - extending Eloquent where clauses depending on dynamic parameters

Answer 2

您可以concat执行此操作，以便将基础查询的动态组合在一起：

$filtered_table = Method::leftJoin('users', 'users.id', '=', 'methods.created_by')
      ->leftJoin('roles', 'roles.id', '=', 'users.role_id')
      ->leftJoin('types', 'types.id', '=', 'methods.type_id');

      foreach ($request->filters as $key => $value) {
        $filtered_table .= $filtered_table->where($key, '=', $value);
      }//dynamic where parameters use `concat` to make the query

      $filtered_table->get([ 'users.username', 'users.id AS users_id', 'methods.*', 'methods.id AS method_id', 'methods.name AS method_name', 'roles.id AS role_id', 'roles.name AS role_name',
      'types.id AS type_id_typetable', 'types.name AS type_name']);

动态参数laravel

2 个答案: