我有一个非常基本的声明，我正在尝试运行，我有一个问题。
$item = $this->db
->select("r.CustomerIDs, r.DateAdded")
->join("customer_orders_rewards as cor", "r.RewardID = cor.RewardID")
->join("customer_orders as co", "co.OrderID = cor.OrderID")
->where(array("r.Denomination" => $row['Denomination'], "r.RewardID" => "cor.RewardID"))
->get("customer_rewards as r");在上面的陈述中，它是解释 cor.RewardID 作为字符串，我希望它是 join .
它导致查询如下所示：
SELECT `r`.`CustomerID`, `r`.`DateAdded`
FROM `customer_rewards` as `r`
JOIN `customer_orders_rewards` as `cor`
ON `r`.`RewardID` = `cor`.`RewardID`
JOIN `customer_orders` as `co`
ON `co`.`OrderID` = `cor`.`OrderID`
WHERE `r`.`Denomination` = '35'
AND `r`.`RewardID` = 'cor.RewardID' <---- Issue如何从 join 在我的 WHERE 条款？
1条答案
按热度按时间wj8zmpe11#
一个简单的解决方法是
而不是
那样的话，
cor.RewardID不应将其视为字符串文字，而应视为实际列。另一方面，你已经加入行时
r.RewardID = cor.RewardID，所以我想说，额外的条件是多余的（有问题的一个），使它不需要。