我有主题(id *)和标签(id *,名称)和链接表topic_tags(topicFk *,tagFk *)。
现在我想选择每个主题,包含所有好标签(a,b,c),但没有坏标签(d,e,f)。
我该怎么做?
答案 0 :(得分:5)
假设您的Topic_Tags表格是唯一的,这会回答您的完全问题 - 但可能无法解决您的实际问题:
SELECT
TopicId
FROM Topic_Tags
JOIN Tags ON
Topic_Tags.TagId = Tags.TagId
WHERE
Tags.Name IN ('A', 'B', 'C', 'D', 'E', 'F')
GROUP BY
TopicId
HAVING
COUNT(*) = 3
AND MAX(Tags.Name) = 'C'
更通用的解决方案是:
SELECT
*
FROM (
SELECT
TopicId
FROM Topic_Tags
JOIN Tags ON
Topic_Tags.TagId = Tags.TagId
WHERE
Tags.Name IN ('A', 'B', 'C')
GROUP BY
TopicId
HAVING
COUNT(*) = 3
) as GoodTags
LEFT JOIN (
SELECT
TopicId
FROM Topic_Tags
JOIN Tags ON
Topic_Tags.TagId = Tags.TagId
WHERE
Tags.Name = 'D'
OR Tags.Name = 'E'
OR Tags.Name = 'F'
) as BadTags ON
GoodTags.TopicId = BadTags.TopicId
WHERE
BadTags.TopicId IS NULL
答案 1 :(得分:3)
这是一个可行的解决方案,但需要为您需要的每个标记提供连接。
SELECT * FROM topics WHERE topic_id IN (SELECT topic_id FROM topic_tags a INNER JOIN topic_tags b on a.topic_id=b.topic_id and b.tag = 'b' INNER JOIN topic_tags c on b.topic_id=c.topic_d and c.tag = 'c' WHERE a.tag = 'a') AND topic_id NOT IN (SELECT topic_id FROM topic_tags WHERE tag = 'd' or tag = 'e' or tag = 'f')
答案 2 :(得分:3)
这是另一个替代查询。也许将好的和坏的标签列表放在顶部更清晰,更方便。我在MySQL 5.0上测试了这个。
SELECT t.*,
SUM(CASE WHEN g.name IN ('a', 'b', 'c') THEN 1 ELSE 0 END) AS num_good_tags,
SUM(CASE WHEN g.name IN ('d', 'e', 'f') THEN 1 ELSE 0 END) AS num_bad_tags
FROM topics AS t
JOIN topic_tags AS tg ON (t.id = tg.topicFk)
JOIN tags AS g ON (g.id = tg.tagFk)
GROUP BY t.id
HAVING num_good_tags = 3 AND num_bad_tags = 0;
答案 3 :(得分:1)
正如所写,其他3个答案进来了,但这是不同的所以我会发布它。
我们的想法是选择包含a,b,c标签的所有主题,然后使用左连接标识那些也包含d,e,f的主题,然后使用where子句过滤那些主题,并在其上查找空值加入...
select distinct topics.id from topics
inner join topic_tags as t1
on (t1.topicFK=topics.id)
inner join tags as goodtags
on(goodtags.id=t1.tagFK and goodtags.name in ('a', 'b', 'c'))
left join topic_tags as t2
on (t2.topicFK=topics.id)
left join tags as badtags
on(badtags .id=t2.tagFK and batags.name in ('d', 'e', 'f'))
where badtags.name is null;
完全没有经过测试,但希望你能看到逻辑的来源。
答案 4 :(得分:0)
我不完全确定,我希望有更好的方法来做好标签部分,但是:
select id from topic
inner join topic_tags tta on topic.id=tta.topicFk and tta.tagFk=a
inner join topic_tags ttb on topic.id=ttb.topicFk and ttb.tagFk=b
inner join topic_tags ttc on topic.id=ttc.topicFk and ttc.tagFk=c
left join topic_tags tt on topic.id=tt.topicFk and tt.tagFk in (d,e,f)
where tt.topicFk is null;
更新:这样的事情:
select id from topic
left join topic_tags tt on topic.id=tt.topicFk and tt.tagFk in (d,e,f)
where tt.topicFk is null and
3=(select count(*) from topic_tags where topicFk=topic.id and tagFk in (a,b,c));
我看到一个答案,假设a,b,c,d,e,f是名称,而不是ID。如果是这样,那么:
select id from topic
left join topic_tags tt on topic.id=tt.topicFk
inner join tags on tt.tagFk=tags.id and tags.name in (d,e,f)
where tt.topicFk is null and
3=(select count(*) from tags inner join topic_tags on tags.id=topic_tags.tagFk and topic_tags.topicFk=topic.id where tags.name in (a,b,c));
答案 5 :(得分:0)
您可以使用minus
关键字来过滤掉有不良标记的主题。
-- All topics with desired tags.
select distinct T.*
from Topics T inner join Topics_Tags R on T.id = R.topicFK
inner join Tags U on U.id = R.topic=FK
where U.name in ('a', 'b', 'c')
minus
-- All topics with undesired tags. These are filtered out.
select distinct T.*
from Topics T inner join Topics_Tags R on T.id = R.topicFK
inner join Tags U on U.id = R.topic=FK
where U.name in ('d', 'e', 'f')
答案 6 :(得分:0)
我自己的解决方案使用Pauls and Bills的想法。
我们的想法是使用好的标签来内连接主题(抛弃没有好标签的主题),然后计算每个主题的唯一标签(以验证所有好的标签都存在)。
同时,带有错误标记的外部联接应该没有一个匹配项(所有字段都是NULL)。
SELECT topics.id
FROM topics
INNER JOIN topic_tags topic_ptags
ON topics.id = topic_ptags.topicFk
INNER JOIN tags ptags
ON topic_ptags.tagFk = ptags.id
AND ptags.name IN ('a','b','c')
LEFT JOIN topic_tags topic_ntags
ON topics.id = topic_ntags.topicFk
LEFT JOIN tags ntags
ON topic_ntags.tagFk = ntags.id
AND ntags.name IN ('d','e','f')
GROUP BY topics.id
HAVING count(DISTINCT ptags.id) = 3
AND count(ntags.id) = 0