我的示例数据如下所示,我的问题是如何根据提供的输入电子邮件查找朋友的电子邮件
| users |
| user_id | email |
|---------|--------------------------|
| 1 | ashutosh8657@example.com |
| 2 | Kanchhi@example.com |
| 3 | modi@example.com |
| 4 | andy@example.com |
| 5 | maya@example.com |
| 6 | jetli@example.com |
| 7 | john@example.com |
| user_relations |
| user_relation_id | requestor_user_id | receiver_user_id | friend_status |
|------------------|-------------------|------------------|---------------|
| 1 | 2 | 4 | 1 |
| 2 | 2 | 6 | 1 |
| 3 | 2 | 7 | 1 |
| 4 | 5 | 2 | NULL |
| 5 | 5 | 7 | NULL |
| 6 | 7 | 2 | NULL |
| 7 | 7 | 4 | 1 |
| 8 | 7 | 5 | 1 |
| 9 | 7 | 6 | 1 |
| 10 | 4 | 2 | 1 |
| 11 | 4 | 3 | 1 |
| 12 | 4 | 5 | 1 |
| 13 | 4 | 6 | 1 |
| 14 | 4 | 7 | 1 |
例1:假设,如果我提供以下输入作为电子邮件:
john@example.com
然后我的预期输出应该是这个(顺序无关紧要):
andy@example.com
jetli@example.com
Kanchhi@example.com
maya@example.com
这里,在user_relations表john@example.com中,其userId为7,对于userId 7,你可以看到朋友是UserIds 4,6,2,5,因此我需要4,6,2,5的电子邮件地址
如果friend_status的值为1,则只考虑朋友,除了1之外他们不是恶魔.我尝试过这个查询,但结果为零.请帮忙
SELECT
case when u.user_id = r.receiver_user_id then requestor.email else receiver.email end as friend_email
FROM users u
JOIN user_relations r
ON (r.requestor_user_id = u.user_id OR r.receiver_user_id = u.user_id)
AND r.friend_status = 1
LEFT JOIN users requestor ON requestor.user_id = r.requestor_user_id
LEFT JOIN users receiver ON receiver.user_id = r.receiver_user_id
WHERE u.email in ('john@example.com')
GROUP BY friend_email
HAVING COUNT(DISTINCT u.user_id) > 1
最佳答案 你正在做2次加入是没用的,这里是你应该如何重写你的查询,在我看来,你不关心朋友是接收者还是请求者:
SELECT DISTINCT
u2.email as friend_email
FROM users u
INNER JOIN user_relations r
ON (r.requestor_user_id = u.user_id OR r.receiver_user_id = u.user_id)
INNER JOIN users u2
ON (r.requestor_user_id = u2.user_id OR r.receiver_user_id = u2.user_id)
AND u2.user_id <> u.user_id
WHERE u.email ='andy@example.com'
AND r.friend_status = 1