Question

我想在laravel 5.2中获得与mysql相同的结果：

select a3.user_id,sum(a3.task_hour)
    from activities a3
    where id not in
    ( select a1.id from activities a1 inner join
    (SELECT * FROM `activities` where validated = 1) a2
    on a1.user_id = a2.user_id and a1.project_id = a2.project_id where a1.validated = 0)
     group by a3.user_id

因此，为了清楚地了解代码，我首先尝试获得以下内容：

(SELECT * FROM `activities` where validated = 1) a2

用这个

$activity_onlyFromOTL = DB::table('activities AS a2');
$activity_onlyFromOTL->select('a2.id','a2.year','a2.month','a2.user_id','a2.project_id','a2.task_hour')->where('a2.from_otl','=','1');

对于第二部分，我有这个：

( select a1.id from activities a1 inner join
        (SELECT * FROM `activities` where validated = 1) a2
        on a1.user_id = a2.user_id and a1.project_id = a2.project_id where a1.validated = 0)

为了清楚起见，我尝试将第一个变量用于第二个语句，以便我可以更好地阅读它：

$activity_whereOTLandNonOTL = DB::table('activities AS a1');
$activity_whereOTLandNonOTL->select('a1.id');
$activity_whereOTLandNonOTL->join($activity_onlyFromOTL, function ($join) {
      $join->on('a1.user_id', '=', 'a2.user_id')->on('a1.project_id', '=', 'a2.project_id');
    });
$activity_whereOTLandNonOTL->where('a1.validated', '=' , '0');

但是当我尝试为那个获取toSql时，我收到错误：

ErrorException in Grammar.php line 39:
Object of class Illuminate\Database\Query\Builder could not be converted to string

我如何以这种方式工作，以便它是一个更清晰的代码？

Laravel使用表的结果作为另一个选择的连接

0 个答案: