这个问题是this one的继续。
我有下表egr
:
+---------+------------+
| offid | groupid |
+---------+------------+
| 1 | 101 |
| 1 | 202 |
| 2 | 202 |
| 2 | 404 |
+---------+------------+
我想插入offid 2所没有的缺少的groupid(与offid 1相比)。结果将是:
+---------+------------+
| offid | groupid |
+---------+------------+
| 1 | 101 |
| 1 | 202 |
| 2 | 202 |
| 2 | 404 |
| 2 | 101 | --> new row to insert
+---------+------------+
我的尝试,基于我其他问题的答案(无效):
INSERT INTO egr (offid, groupid)
SELECT 2, egr1.groupid
FROM egr AS egr1
WHERE egr1.offid = 1
AND NOT EXISTS
(select 1
from egr e2
where e2.groupid = egr1.groupid and
e2.offid in (1, 2) and
e2.offid <> egr1.offid
);
答案 0 :(得分:2)
使用except
运算符可能更容易实现:
INSERT INTO egr (offid, groupid)
SELECT 2, groupid
FROM egr
WHERE offid = 1
EXCEPT
SELECT 2, groupid
FROM egr
WHERE offid = 2