Question

我对SQL并不擅长，但我尽我所能，因为我所拥有的一点点知识......

我在朋友的帮助下创建了一个单独的SQL字符串，该朋友使用来自第一个表的初始数据的相对表从表中收集数据，SQL就像这样：

SELECT id, username, auth_assignment.created_at
FROM `user` JOIN `auth_assignment`
ON (user.id = auth_assignment.user_id)
JOIN `auth_item`
ON (auth_item.name = auth_assignment.item_name)
WHERE auth_item.name = 'Admin'

要查看的初始数据是Admin所以一切都在那边工作，但我尝试使用SQL函数模拟这个Yii2 ..到目前为止，我已经制作了这段代码

$query = new Query;

$command = $query->select('id, username')
    ->innerJoin('auth_assignment', ['user.id'=>'auth_assignment.user_id'])
    ->innerJoin('auth_item', ['auth_item.name'=>'auth_assignment.item_name'])
    ->where('auth_item.name = :name', [':name'=>$name])->createCommand();

var_dump($command->query());

这会返回SQLSTATE错误：

SQLSTATE[42000]: Syntax error or access violation: 1064 You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'INNER JOIN `auth_assignment` ON `user`.`id`='auth_assignment.user_id' INNER JOIN' at line 1
The SQL being executed was: SELECT `id`, `username` INNER JOIN `auth_assignment` ON `user`.`id`='auth_assignment.user_id' INNER JOIN `auth_item` ON `auth_item`.`name`='auth_assignment.item_name' WHERE auth_item.name = 'Admin'

我检查了方法$command->sql;以了解SQL是如何生成的...但由于我对SQL缺乏了解，我真的不知道如何修复它对yii2 api文档缺乏了解

SQL生成如下：

SELECT `id`, `username` INNER JOIN `auth_assignment` ON `user`.`id`=:qp1 INNER JOIN `auth_item` ON `auth_item`.`name`=:qp2 WHERE auth_item.name = :name

我感谢任何帮助

Answer 1

尝试此查询

$query = (new yii\db\Query())
         ->select('id, username, auth_assignment.created_at')
         ->from('user')
         ->innerJoin('auth_assignment','user.id=auth_assignment.user_id')
         ->innerJoin('auth_item','auth_item.name = auth_assignment.item_name')
         ->where([
               'auth_item.name' => 'Admin'
         ])->all();

Yii2：无法使用innerJoin

1 个答案: