如何在 php 中使用 if else 條件和 Join

Question

我想獲取 $id(dynamic,passing as parameter) 的問題，並想知道這是“ParentbusinessId”還是“businessId”

表名“nps_ques”

id      config_id       question
1       1               Lorem Ipsum
2       1               Lorem Ipsum2
3       1               Lorem Ipsum3

表“nps_config”

id      parentBusinessId        businessId
1       4580                    NULL    
2       0                       2
3       4580                    3

我嘗試使用以下代碼

$this->db->select('nq.id as ques_id,nq.config_id as formId,nq.question,nc.parentBusinessId,nc.businessId)
                ->from('nps_ques nq')
                ->join("nps_config nc", "nc.id=nq.config_id")
                ->where("nq.config_id", $id)
                ->order_by("nq.id", "ASC");

Answer 1

您可以使用 CASE。 請參閱下面的示例：

SELECT 
config.id AS config_id,
questions.question,
case when config.parent_business_id is not null then config.parent_business_id else config.business_id end as business
FROM
questions
    JOIN
config ON questions.config_id = config.id
WHERE
  config.id = 1;