query-builder

I have a Model, Articles, which hasMany Abstracts. I want to load the 10 latest Articles, and for each Article, the Abstract with the highest number of points. My function looks like this: public function getArticles($category, $viewName) { $subArticles = $this->Articles->findByCategory($category)->contain([ 'Abstracts' => function ($q) { return $q ->select(['body', 'points', 'article_id']) ->where(['Abstracts.approved' => true]) ->limit(10) ->order(['Abstracts.points' => 'DESC']); } ]) ->limit(10) ->order(['Articles.created' => 'DESC']) ; $this->set( $viewName . 'Articles', $subArticles ); }

I'd like to get value by the following SQL using Eloquent ORM. - SQL SELECT COUNT(*) FROM (SELECT * FROM abc GROUP BY col1) AS a; Then I considered the following. - Code $sql = Abc::from('abc AS a')->groupBy('col1')->toSql(); $num = Abc::from(\DB::raw($sql))->count(); print $num; I'm looking for a better solution. Please tell me simplest solution. In addition to @delmadord's answer and your comments: Currently there is no method to create subquery in FROM clause, so you need to manually use raw statement, then, if necessary, you will merge all the bindings: $sub = Abc::where(..)->groupBy(..);

I have a belongsToMany association on Users and Contacts. I would like to find the Contacts of the given User. I would need something like $this->Contacts->find()->contain(['Users' => ['Users.id' => 1]]); The cookbook speaks about giving conditions to contain, custom finder methods and sing through association key, but I did not find out how to put these together. Use Query::matching() or Query::innerJoinWith() When querying from the Contacts table, then what you are looking for is Query::matching() or Query::innerJoinWith() , not (only) Query::contain() . See Cookbook > Database Access & ORM