我有3个表格，其中包含以下数据：
第一个表称为连接，其中connections.username1是跟随者，connections.username2是被关注的人。
它有以下几行：
connections.username1 | connections.username2
mikha | guy
guy | maricela
maricela | guy
questions.asker_username，另一个用于收到名为questions.target_username的提问者。当提问者被称为“sys.tem”并且目标被称为“every.one”时，它被认为是一个全局问题，可以由所有成员回答。
匿名用户可以询问并将其IP记录为asker_username。
它有以下几行：
questions.id | questions.asker_username | questions.target_username | questions.question
1 | mikha | guy | what's your name?
2 | mikha | maricela | What's your age?
3 | guy | mikha | what's your name?
4 | maricela | guy | favorite food?
5 | xx.xx.xxx.xx | mikha | favorite pet?
6 | xx.xx.xxx.xx | guy | first name?
7 | xx.xx.xxx.xx | maricela | first name?
8 | sys.tem | every.one | what's ur name?
9 | sys.tem | every.one | favorite movie?
10 | sys.tem | every.one | favorite game?
第3个表称为答案。答案表中的id与问题ID相同。此表有一个id，用户名和答案列。
answers.id | answers.username | answers.answer
1 | guy | my name is guy
2 | maricela | my name is maricela
3 | mikha | my name is mikha
4 | guy | pizza
8 | guy | guy is my name
8 | maricela | maricela is my name
9 | maricela | avatar
我想要一个查询，它结合了以下与“mikha”相关的条件和他所关注的人：
1）questions.asker_username不是“mikha”
2）questions.target_username是“mikha”或他跟随的任何用户。
3）如果questions.target_username等于“every.one”并由“mikha”回答，请显示问题。
4）如果questions.target_username等于“every.one”并由“mikha”跟随的任何人回答，请显示问题及其答案。如果“mikha”跟随的用户没有回答，请不要显示问题。
5）如果questions.target_username等于“every.one”并且根本没有任何人回答，请显示一次问题。
6）如果questions.target_username等于“every.one”并且未被“mikha”回答并且没有被他所关注的任何人回答，则只显示一次问题。
我使用以下查询：
SELECT questions.id,answers.id,questions.asker_username,questions.target_username,
answers.username,questions.question,answers.answer
FROM questions
LEFT JOIN answers ON (questions.id = answers.id)
LEFT JOIN connections ON connections.username1 = 'mikha'
AND (questions.target_username = connections.username2
OR questions.asker_username = connections.username2
OR connections.username2 = answers.username)
WHERE questions.asker_username <> 'mikha'
AND (questions.target_username = 'mikha'
OR questions.target_username = connections.username2
OR (questions.target_username = 'every.one'
AND (answers.username = 'mikha'
OR answers.username = connections.username2
OR answers.username IS NULL)
)
)
GROUP BY questions.id,answers.username
我期待的结果：
questions.id | answers.id | questions.asker_username | questions.target_username | answers.username | questions.question | answers.answer
3 | 3 | guy | mikha | mikha | what's your name? | my name is mikha
4 | 4 | maricela | guy | guy | favorite food? | pizza
5 | 5 | xx.xx.xxx.xx | mikha | NULL | favorite pet? | NULL
6 | 6 | xx.xx.xxx.xx | guy | NULL | first name? | NULL
8 | 8 | sys.tem | every.one | NULL | what's ur name? | NULL
8 | 8 | sys.tem | every.one | guy | what's ur name? | guy is my name
9 | 9 | sys.tem | every.one | NULL | favorite movie? | NULL
10 | 10 | sys.tem | every.one | NULL | favorite game? | NULL
我实际得到的结果：
questions.id | answers.id | questions.asker_username | questions.target_username | answers.username | questions.question | answers.answer
3 | 3 | guy | mikha | mikha | what's your name? | my name is mikha
4 | 4 | maricela | guy | guy | favorite food? | pizza
5 | 5 | xx.xx.xxx.xx | mikha | NULL | favorite pet? | NULL
6 | 6 | xx.xx.xxx.xx | guy | NULL | first name? | NULL
8 | 8 | sys.tem | every.one | guy | what's ur name? | guy is my name
10 | 10 | sys.tem | every.one | NULL | favorite game? | NULL
我在http://sqlfiddle.com/#!2/29929e/1上建立了一个方案，向您展示我实际得到的结果
谢谢：）
答案 0 :(得分：5)
好的，让我们从最简单的（你的第一条规则）开始：
SELECT q.id, a.id, q.asker_username, q.target_username, a.username, q.question, a.answer
FROM questions q
LEFT JOIN answers a ON q.id = a.id
WHERE q.asker_username <> 'mikha'
GROUP BY q.id,a.username
现在让我们添加你的第二条规则 - 现在增加了更多的复杂性......
SELECT q.id, a.id, q.asker_username, q.target_username, a.username, q.question, a.answer
FROM questions q
LEFT JOIN answers a ON q.id = a.id
WHERE q.asker_username <> 'mikha'
AND q.target_username = 'mikha'
OR q.target_username IN (
SELECT username2
FROM connections
WHERE username1 = 'mikha'
)
GROUP BY q.id,a.username
现在第三条规则（由mikha回答的所有人）：
SELECT q.id, a.id, q.asker_username, q.target_username, a.username, q.question, a.answer
FROM questions q
LEFT JOIN answers a ON q.id = a.id
WHERE q.asker_username <> 'mikha'
AND q.target_username = 'mikha'
OR q.target_username IN (
SELECT username2
FROM connections
WHERE username1 = 'mikha'
)
OR (q.target_username = 'every.one' AND a.username = 'mikha')
GROUP BY q.id,a.username
现在是第四条规则：
SELECT q.id, a.id, q.asker_username, q.target_username, a.username, q.question, a.answer
FROM questions q
LEFT JOIN answers a ON q.id = a.id
WHERE q.asker_username <> 'mikha'
AND q.target_username = 'mikha'
OR q.target_username IN (
SELECT username2
FROM connections
WHERE username1 = 'mikha'
)
OR (q.target_username = 'every.one' AND a.username = 'mikha')
OR (q.target_username = 'every.one' AND a.username IN (
SELECT username2
FROM connections
INNER JOIN answers ON answers.username = connections.username2
AND answers.answers IS NOT NULL
WHERE username1 = 'mikha'
))
GROUP BY q.id,a.username
第五条规则（耶稣！）：
SELECT q.id, a.id, q.asker_username, q.target_username, a.username, q.question, a.answer
FROM questions q
LEFT JOIN answers a ON q.id = a.id
WHERE q.asker_username <> 'mikha'
AND q.target_username = 'mikha'
OR q.target_username IN (
SELECT username2
FROM connections
WHERE username1 = 'mikha'
)
OR (q.target_username = 'every.one' AND a.username = 'mikha')
OR (q.target_username = 'every.one' AND a.username IN (
SELECT username2
FROM connections
INNER JOIN answers ON answers.username = connections.username2
AND answers.answers IS NOT NULL
WHERE username1 = 'mikha'
))
OR (q.target_username = 'every.one' AND a.answer IS NULL)
GROUP BY q.id,a.username
最后一个：
SELECT q.id, a.id, q.asker_username, q.target_username, a.username, q.question, a.answer
FROM questions q
LEFT JOIN answers a ON q.id = a.id
WHERE q.asker_username <> 'mikha'
AND q.target_username = 'mikha'
OR q.target_username IN (
SELECT username2
FROM connections
WHERE username1 = 'mikha'
)
OR (q.target_username = 'every.one' AND a.username = 'mikha')
OR (q.target_username = 'every.one' AND a.username IN (
SELECT username2
FROM connections
INNER JOIN answers ON answers.username = connections.username2
AND answers.answers IS NOT NULL
WHERE username1 = 'mikha'
))
OR (q.target_username = 'every.one' AND a.answer IS NULL)
OR (q.target_username = 'every.one' AND a.username NOT IN (
SELECT username2
FROM connections
INNER JOIN answers ON answers.username = connections.username2
AND answers.answers IS NOT NULL
WHERE username1 = 'mikha'
))
GROUP BY q.id,a.username
我认为规则4和规则6有点相互对立（可以说是矛盾的），当在一个查询中使用时，它会产生与省略相同的效果......
我没有测试任何查询，但我相信它们有效。
答案 1 :(得分：2)
问题在于 - 在某些情况下 - 当只有一个匹配的答案存在时，您想要显示两次问题。我引用：
所以，我想只为“mikha”展示一次它是否得到回答，并且每当任何人回答“mikha”后再次显示它
这种重复使得事情变得非常困难。
我尝试使用UNION解决这个问题，但它似乎有效。但是，我仍然没有完全理解你的要求......
无论如何，我们走了：
select * from
(
select
q.id as q_id, a.id as a_id, q.asker_username,
q.target_username, a.username, q.question, a.answer
from
questions q
left outer join answers a on q.id = a.id
where
q.asker_username <> 'mikha'
and
(
q.target_username = 'mikha'
or q.target_username in
(select username2 from connections where username1 = 'mikha')
or
(
q.target_username = 'every.one'
and
(
a.username = 'mikha'
or a.username in
(select username2 from connections where username1 = 'mikha')
or a.id is null
)
)
)
union
select
q.id as q_id, NULL as a_id, q.asker_username,
q.target_username, NULL, q.question, NULL
from
questions q
where
q.asker_username <> 'mikha'
and q.target_username = 'every.one'
and not exists (select id
from answers
where
id = q.id
and username = 'mikha'
)
) r
order by q_id;
现场测试：
With answer from mikha for question 8
Without answer from mikha for question 8
答案 2 :(得分：0)
我建议你的起点之一是你需要加入你的连接表两次 - 一次是答案，一次是问题。
下面的代码给出了它正在做什么的内联评论。我也同意其他答案，建议数字id字段更适合比较 - 我还建议在答案表上放置一个唯一的rowid字段（更好的是，将id更改为question_id并且然后将id设为唯一字段。
SELECT DISTINCT q.id,a.id,q.asker_username,q.target_username,a.username,q.question,a.answer
FROM questions q
/* Answers */
LEFT JOIN answers a ON (q.id=a.id)
/* connection entries where the person being followed is the target */
LEFT JOIN (SELECT username2 FROM connections WHERE username1='mikha') c_q
ON c_q.username2=q.target_username
/* connection entries where the person being followed answered the question */
LEFT JOIN (SELECT username2 FROM connections WHERE username1='mikha') c_a
ON c_a.username2=a.username
/* Own answers */
LEFT JOIN (SELECT id FROM answers WHERE username='mikha') a_own
ON (q.id=a_own.id)
WHERE
/* Asker not mikha, target is mikha or followed user - rules 1,2 */
(q.asker_username <> 'mikha'
AND (q.target_username='mikha' OR c_q.username2 IS NOT NULL))
OR
/* sys.tem/every.one, answered by mikha - rule 3 */
(q.target_username='every.one' AND a.username='mikha')
OR
/* Rules 4, 5 & 6 combine to give "show the answer at least once, and once for every followed user who answered" - here we select any every.one messages where mikha didn't answer and nor did the people he follows */
(q.target_username='every.one' AND a_own.id IS NULL AND c_a.username2 IS NULL);
对于问题8，它没有给你两行，但我无法确定你为什么要这样做。问题在于，不清楚哪些规则是加法的，哪些是组合的。如果您可以解释哪些规则应该“添加”一行而不仅仅是提供另一个显示行的原因，那么我们也可以为问题8获取额外的行。