【发布时间】:2020-03-20 01:40:53
【问题描述】:
我在 Laravel 中有这个查询:
DB::table('score')
->select('score.score_nl', DB::raw('count(*) as total'), DB::raw('round(avg(rating_results.rating)) as final_rating'))
->join('rating', 'rating.score_id', '=', 'score.id')
->join('rating_results', 'rating.rating_result_id', '=', 'rating_results.id')
->groupBy('score_nl')
->get();
结果:
[{"score_nl":"emphatisch","total":1,"final_rating":"1"},{"score_nl":"huilen","total":2,"final_rating":"3"},{"score_nl":"knuffelig","total":1,"final_rating":"1"},{"score_nl":"zindelijkheid","total":2,"final_rating":"3"}]
我想在这个表中有一个名为rating_results(见图片)的表
查找final_rating 并获取关联的result_en。
我怎么能在 laravel 中做到这一点?
有任何问题请告诉我!
--编辑--
这个我试过了;
$q = Result::select('result_nl')
->whereColumn('rating_results.rating', 'final_rating')
->whereColumn('rating_results.score_id', 'score_id')
->getQuery();
DB::table('score')
->select('score.score_nl', DB::raw('count(*) as total'), DB::raw('round(avg(rating_results.rating)) as final_rating'))
->join('rating', 'rating.score_id', '=', 'score.id')
->join('rating_results', 'rating.rating_result_id', '=', 'rating_results.id')
->selectSub($q, 'result_nl')
->groupBy('score_nl')
->get();
然后我得到这个错误:
SQLSTATE[42S22]: Column not found: 1247 Reference 'final_rating' not supported (reference to group function) (SQL: select `score`.`score_nl`, count(*) as total, round(avg(rating_results.rating)) as final_rating, (select `result_nl` from `rating_results` where `rating_results`.`rating` = `final_rating` and `rating_results`.`score_id` = `score_id`) as `result_nl` from `score` inner join `rating` on `rating`.`score_id` = `score`.`id` inner join `rating_results` on `rating`.`rating_result_id` = `rating_results`.`id` group by `score_nl`)
看来我需要使用joinSub
【问题讨论】:
标签: mysql sql laravel laravel-query-builder