Php 将codeigniter查询转换为sql查询

在编写SQL查询以连接多个表时，我有点困惑。 我已经在codeigniter中完美地编写了以下查询

$this->db->select('*');
$this->db->from('subscription_payment'); 
$this->db->join('user', 'user.user_id = subscription_payment.user_id');
$this->db->join('subscription', 'subscription.subscription_id = 
subscription_payment.subscription_id');
$this->db->where('subscription_payment.subscription_payment_id',$sid);
$query = $this->db->get(); 
$result= $query->result_array();

请帮助我使用将此转换为SQL查询

SELECT *
FROM subscription_payment
JOIN user ON user.user_id = subscription_payment.user_id
JOIN subscription ON subscription.subscription_id = subscription_payment.subscription_id
WHERE subscription_payment.subscription_payment_id = $sid 
;

将产生

select * from some_table...

---

就是这样。下次您可以使用easy转换任何查询。

注意：如果您无法生成最后一个查询，请尝试检查数据库保存查询是否设置为true，默认情况下是否为true

echo $this->db->last_query();

select * from some_table...