左连接返回所有结果，即使条件不满足

Question

我有一个简单的查询

select * 
from employees as e 
left join email_notification as en 
on e.id=en.eid and en.`status` != 'SUCCESS';

现在我在 email_notification table 中有一个记录状态为成功，但仍然从员工 table 获取该记录。

Answer 1

我猜你的意思是：

select *
from employees as e
left join email_notification as en
on e.id=en.eid
where en.`status` != 'SUCCESS';

我的意思是，status 的检查不应与 join 一起进行，而是参与最终结果。

Answer 2

您需要将加入员工加入 "successes" 并选择那些没有加入的员工，如下所示：

SELECT e.* 
FROM employees AS e 
LEFT email_notification AS en 
ON e.id=en.eid and en.`status` = 'SUCCESS'
WHERE en.id IS NULL
;

Answer 3

为什么不尝试使用 IN？

select * 
from employees as e 
where id not in(
    select eid from email_notification 
    where status != 'SUCCESS'
)

Left join returning all the results even when the condition is not satisfied