Php 如何在JOIN语句中使用CASE根据条件连接两个表？

我有三张表格：电子邮件、新客户和旧客户。 现在在email表中有一个名为type的字段，其中包含0或1
因此，如果type包含0，那么我想从customer\u new表中获取name列，从customer\u old表中获取其他列

为此，我进行了以下mysql查询：

SELECT email.*, 
CASE email.`type`
    WHEN 1 THEN customer_old.name
    ELSE customer_new.name
END AS customer_name 
LEFT JOIN 
CASE email.`type`
    WHEN 0 THEN customer_new ON email.customer_id = customer_new.id
    ELSE customer_old ON email.customer_id = customer_old.id
END
FROM email

现在，当我运行这个查询时，mysql总是抛出错误。
有人能根据条件通过CASE语句解决我关于连接两个表的问题吗

提前感谢。

尝试下面的查询

SELECT email . * , IF( email.type = '1', customer_old.name, customer_new.name ) AS name 
FROM email
LEFT JOIN customer_old ON ( customer_old.id = email.customer_id ) 
LEFT JOIN customer_new ON ( customer_new.id = email.customer_id )

您可以使用left join优化执行。

另一种选择

SELECT * FROM email 
LEFT JOIN customer_new AS customer ON email.customer_id = customer.id 
WHERE email.type = 0

UNION

SELECT * FROM email 
LEFT JOIN customer_old AS customer ON email.customer_id = customer.id 
WHERE email.type = 1

---

你能举一个使用它的例子吗？

SELECT * FROM email 
LEFT JOIN customer_new AS customer ON email.customer_id = customer.id 
WHERE email.type = 0

UNION

SELECT * FROM email 
LEFT JOIN customer_old AS customer ON email.customer_id = customer.id 
WHERE email.type = 1