如何組由yii2

Question

我想生成下面的查詢使用yii2計數和：

SELECT COUNT(*) AS cnt FROM lead WHERE approved = 1 GROUP BY promoter_location_id, lead_type_id

我曾嘗試：

$leadsCount = Lead::find() 
->where('approved = 1') 
->groupBy(['promoter_location_id', 'lead_type_id']) 
->count(); 

生成此查詢：

SELECT COUNT(*) FROM (SELECT * FROM `lead` WHERE approved = 1 GROUP BY `promoter_location_id`, `lead_type_id`) `c` 

在yii 1.x我會做以下事情：

$criteria = new CDbCriteria(); 
$criteria->select = 'COUNT(*) AS cnt'; 
$criteria->group = array('promoter_location_id', 'lead_type_id'); 

謝謝！

Answer 1

解決方案：

$leadsCount = Lead::find() 
->select(['COUNT(*) AS cnt']) 
->where('approved = 1') 
->groupBy(['promoter_location_id', 'lead_type_id']) 
->all(); 

，並添加public $cnt這個模型，在我的情況的鉛。

正如Kshitiz所述，您也可以使用yii\db\Query::createCommand()。

Answer 2

如果您只是對計數感興趣，請使用其他人提到的yii\db\Query。不會要求您模型中的任何改變：

$leadsCount = (new yii\db\Query()) 
    ->from('lead') 
    ->where('approved = 1') 
    ->groupBy(['promoter_location_id', 'lead_type_id']) 
    ->count(); 

這裏給Yii2 API documentation

Answer 3

一個鏈接，您可以通過在SELECT查詢

$leadCount = Lead::find() 
->where(['approved'=>'1']) 
->groupBy(['promoter_location_id', 'lead_type_id']) 
->count(); 

使用計數（）得到計數Reference Link用於選擇查詢的各種功能

Answer 4

不添加$cnt原型rty to model

$leadsCount = Lead::find() 
->select(['promoter_location_id', 'lead_type_id','COUNT(*) AS cnt']) 
->where('approved = 1') 
->groupBy(['promoter_location_id', 'lead_type_id']) 
->createCommand()->queryAll();