如何将子查询作为表名传递给yii2中的另一个查询 [英] How to pass subquery as table name to another query in yii2

问题描述

我有一个查询，试图将其转换为yii2语法.下面是查询

I have a query which I am trying to convert into yii2 syntax. Below is the query

 SELECT project_id, user_ref_id FROM 
            (
            SELECT `project_id`, `user_ref_id`
            FROM `projectsList` 
            WHERE user_type_ref_id = 1) AS a WHERE user_ref_id = '.yii::$app->user->id;

我正在尝试将其转换为yii2格式，例如

I am trying to convert it into yii2 format like

 $subQuery = (new Query())->select(['p.project_id', 'p.user_ref_id'])->from('projectsList')->where(['user_type_ref_id' => 1]);

 $uQuery = (new Query())->select(['p.project_id', 'p.user_ref_id'])->from($subQuery)->where(['user_ref_id ' => yii::$app->user->id])->all();

出现类似

 trim() expects parameter 1 to be string, object given

如何将子查询作为表名传递给另一个查询

How to I pass subquery as table name to another query

推荐答案

未经测试，但通常是这样.您需要将subQuery作为表传递.因此，将第二个查询中的->from($subQuery)更改为->from(['subQuery' => $subQuery])

Not tested, but generally this is how it goes. You need to pass the subQuery as a table. So change ->from($subQuery) in the second query to ->from(['subQuery' => $subQuery])

$subQuery = (new Query())->select(['p.project_id', 'p.user_ref_id'])->from('projectsList')->where(['user_type_ref_id' => 1]);

然后

 $query = (new Query())->select(['p.project_id', 'p.user_ref_id'])->from(['subQuery' => $subQuery])->where(['subQuery.user_ref_id ' => yii::$app->user->id])->all();

这篇关于如何将子查询作为表名传递给yii2中的另一个查询的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！