Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
390 views
in Technique[技术] by (71.8m points)

sql - Retrieve 2 last posts for each category

Lets say I have 2 tables: blog_posts and categories. Each blog post belongs to only ONE category, so there is basically a foreign key between the 2 tables here.

I would like to retrieve the 2 lasts posts from each category, is it possible to achieve this in a single request? GROUP BY would group everything and leave me with only one row in each category. But I want 2 of them.

It would be easy to perform 1 + N query (N = number of category). First retrieve the categories. And then retrieve 2 posts from each category.

I believe it would also be quite easy to perform M queries (M = number of posts I want from each category). First query selects the first post for each category (with a group by). Second query retrieves the second post for each category. etc.

I'm just wondering if someone has a better solution for this. I don't really mind doing 1+N queries for that, but for curiosity and general SQL knowledge, it would be appreciated!

Thanks in advance to whom can help me with this.

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

Check out this MySQL article on how to work with the top N things in arbitrarily complex groupings; it's good stuff. You can try this:

SET @counter = 0;
SET @category = '';

SELECT
  *
FROM
(
  SELECT
    @counter := IF(posts.category = @category, @counter + 1, 0) AS counter,
    @category := posts.category,
    posts.*
    FROM
      (
      SELECT
        *
        FROM test
        ORDER BY category, date DESC
      ) posts
) posts
HAVING counter < 2

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...