查询以查找带有一组精确标签的帖子(多对多关系) [英] Query to find posts with an exact set of tags (many-to-many relationship)

问题描述

假设我有以下三个表来表达一种关系，其中帖子被赋予标签(多对多关系):

Suppose I have the following three tables expressing a relationship where posts are given tags (a many-to-many relationship):

create table posts (id integer, content text, primary key (id));
create table tags (tag varchar(30), description text, primary key (tag));
create table post_tags (post_id integer, tag varchar(10),
    primary key (post_id, tag),
    foreign key (post_id) references posts (id),
    foreign key (tag) references tags (tag));

现在假设我想查找所有带有标签{clever，interesting}，而没有其他标签的帖子.

Now suppose I want to find all posts with exactly the tags {clever,interesting}, and no others.

这是我失败的尝试.它会找到带有标签{clever，interesting}的帖子，但还会找到带有标签{clever，interesting，annoying}或{clever，interesting，unthical}的帖子.

Here is my failed attempt. It finds posts which have the tags {clever,interesting}, but it also finds posts with the tags {clever,interesting,annoying} or {clever,interesting,unethical}.

select t1.post_id from post_tags as t1
    inner join post_tags as t2 on t2.post_id=t1.post_id
    where t1.tag='clever' and t2.tag='interesting';

我乐于接受任何有关更改结构以简化此操作的建议.但是，我想避免在posts表中添加逗号分隔的标签列表之类的事情.

I'm open to any suggestions for changing the structure to make this operation easier. However, I want to avoid things like adding a comma-separated list of tags in the posts table.

推荐答案

这是一个精确的关系划分问题.

This is an exact relational division problem.

在SQL Server中，运行良好方法(假设对post_id,tag的唯一约束)为

In SQL Server a well performing method (assuming unique constraint on post_id,tag) is

SELECT post_id
FROM   post_tags
GROUP  BY post_id
HAVING MIN(CASE
             WHEN Keyword IN ( 'clever', 'interesting' ) THEN 1
             ELSE 0
           END) = 1
       AND SUM(CASE
                 WHEN Keyword IN ( 'clever', 'interesting' ) THEN 1
                 ELSE 0
               END) = 2  

所以我不排除在HAVING中使用GROUP_CONCAT的想法.

So I wouldn't rule out the idea of using GROUP_CONCAT in the HAVING instead.

HAVING GROUP_CONCAT(DISTINCT Keyword ORDER BY Keyword) = 'clever,interesting'

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