【发布时间】:2016-08-13 12:30:00
【问题描述】:
select users.*, user_address.*, user_kyc_details.* , user_nominee.*, user_bank_details.*
from users, user_address, user_kyc_details, user_nominee, user_bank_details
where user_address.user_id=users.id and user_kyc_details.user_id=users.id and user_nominee.user_id=users.id and user_bank_details.user_id=users.id
如何在 $query->createCommand() 中直接运行它;作为参数我试过了,但它不起作用
上面的查询工作正常并给出了预期的结果,但是当我使用如下所示的 Yii 加入时,它给了我奇怪的结果(即没有加入适当的 id 和 user_id)
$query = 新查询;
$query->select([
'users.*',
'user_address.*'
'user_kyc_details.*',
'user_nominee.*',
'user_bank_details.*'
])
->from('users')
->join('INNER JOIN', 'user_address','user_address.user_id = users.id')
->join('INNER JOIN', 'user_kyc_details','users.id =user_kyc_details.user_id')
->join('INNER JOIN', 'user_nominee','users.id =user_nominee.user_id')
->join('INNER JOIN', 'user_bank_details','users.id =user_bank_details.user_id');
$command = $query->createCommand();
$data = $command->queryAll();
任何帮助将不胜感激
【问题讨论】:
-
select users.*, user_address.*, user_kyc_details.* , user_nominee.*, user_bank_details.* from users, user_address, user_kyc_details, user_nominee, user_bank_details where user_address.user_id=users.id and user_kyc_details.user_id =users.id 和 user_nominee.user_id=users.id 和 user_bank_details.user_id=users.id
标签: yii inner-join