我有一个非常基本的声明,我试图运行,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");
在上面的语句中,它将
cor.RewardID
解释为一个字符串,我希望它是join
中的一个字段。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
如何引用
join
子句中WHERE
中的字段? 最佳答案
一个简单的解决方法是
"r.RewardID = cor.RewardID"
而不是
"r.RewardID" => "cor.RewardID"
这样,
cor.RewardID
就不应该被视为字符串文字,而应该被视为实际的列。另一方面,当
r.RewardID = cor.RewardID
时,您已经在连接行了,所以我要说额外的条件是多余的(有问题的条件),因此不需要它。