Yii2 innerJoin()

时间:2015-09-09 13:23:23

标签: sql yii2 inner-join

我想以下列方式实现sql查询:

INNER JOIN
`Product_has_ProductFeature` t ON `Product`.`id` = t.`productId` AND t.`productFeatureValueId` = 1
INNER JOIN
`Product_has_ProductFeature` t1 ON `Product`.`id` = t1.`productId` AND t1.`productFeatureValueId` = 5

如何使用innerJoin()或类似的内容完成此操作?

4 个答案:

答案 0 :(得分:12)

您可以使用以下代码:

$model = Product::find()
->innerJoinWith('t', 'Product.id = T.productId')
->andWhere(['T.productFeatureValueId' => ''])
->innerJoinWith('t1', 'Product.id = T1.productId')
->andWhere(['T1.productFeatureValueId' => '5'])
->all();

答案 1 :(得分:4)

innerJoin()是来自method类的Query

你可以尝试这样的事情。

$query = new \yii\db\Query;
$command = $query->innerJoin(
         'Product_has_ProductFeature',
         `Product`.`id` = t.`productId`)
     ->andWhere('t.`productFeatureValueId` = 1')
     ->createCommand();
$queryResult = $command->query();

答案 2 :(得分:2)

$query = new \yii\db\Query();
$query->from(['u' => 'usr_user'])
    ->select(['u.id','p.first_name','p.last_name'])
    ->innerJoin(['i' => 'pdl_inspector'],'`u`.`id` = `inspector_id`')
    ->innerJoin(['p'=>'usr_profile'],'`p`.`user_id` = `u`.`id`')
    ->all();

答案 3 :(得分:1)

表vote_results,vote_answers,vote_questions
关系
1. vote_results> vote_answers(answer_id)
2. vote_answers> vote_questions(question_id)

 $matches = Results::find()
                ->select(['vote_results.id', 'COUNT(vote_results.answer_id) as MatchCount'])
                ->innerJoin('vote_answers', 'vote_results.answer_id = vote_answers.id')
                ->innerJoin('vote_questions', 'vote_answers.question_id = vote_questions.id')
                ->andWhere(['in', 'vote_results.answer_id', 31])
                ->having('COUNT(vote_results.id)>=1')
                ->orderBy('MatchCount DESC')
                ->asArray()
                ->one();
            var_dump($matches['MatchCount']);