MySQL，多行获取一组中一行的position

Question

根据effective_publishing_date _publishing_date，我有这个查询来计算一组出版物（这里称为community ）内的publication的 position ：

SELECT p.publication_id
     , p.name publication_name
     , IF(p.scheduled_at is not null, p.scheduled_at, p.created_at) effective_publishing_date
     , @current_rank := @current_rank + 1 publication_rank
FROM publications p
JOIN (SELECT @current_rank := 0) r 
WHERE p.community_id = 8513
ORDER 
    BY effective_publishing_date ASC;

结果为：

[![在此处输入图像描述][1]][1]

现在我有一个feed_item列表，其中每个community_id和publication_id作为属性，我想为每个feed_item获取相关的publication_rank 。

例如，如果我有一个publication_item = 18 和community_id = 2 的publication_id ，我希望publication_rank #2 community_id所有出版物中publication_id #18 的publication_rank。 我没有成功在一个查询（或子查询等）中得到它。

感谢提前，

Answer 1

这是我最终找到的两个解决方案。 感谢@Barmar！

1. 仅使用连接：

SELECT 
  g1.community_id, 
  g1.publication_name, 
  g1.publication_name, 
  g1.publication_id, 
  COUNT(*) AS rank 
FROM 
  (
    SELECT 
      publications.publication_id as publication_id, 
      publications.name as publication_name, 
      publications.community_id as community_id, 
      communities.name as community_name, 
      IF(
        publications.scheduled_at is not null, 
        publications.scheduled_at, publications.created_at
      ) as effective_publishing_date 
    FROM 
      feed_items 
      JOIN publications ON feed_items.publication_id = publications.publication_id 
      JOIN communities ON publications.community_id = communities.community_id 
    WHERE 
      feed_items.user_id = 489387
  ) AS g1 
  JOIN (
    SELECT 
      publications.publication_id as publication_id, 
      publications.community_id as community_id, 
      IF(
        publications.scheduled_at is not null, 
        publications.scheduled_at, publications.created_at
      ) as effective_publishing_date 
    FROM 
      feed_items 
      JOIN publications ON feed_items.publication_id = publications.publication_id 
    WHERE 
      feed_items.user_id = 489387
  ) AS g2 ON (
    g2.effective_publishing_date, g2.publication_id
  ) <= (
    g1.effective_publishing_date, g1.publication_id
  ) 
  AND g1.community_id = g2.community_id 
GROUP BY 
  g1.publication_id, 
  g1.community_id, 
  g1.effective_publishing_date 
ORDER BY 
  g1.community_id, 
  rank ASC;

2. 带 SQL 变量

SELECT data_table.publication_id, data_table.publication_name, data_table.community_id, data_table.effective_publishing_date,
   @publication := IF(@community <> data_table.community_id, concat(left(@community := data_table.community_id, 0), 0), @publication+1) AS rank
FROM
  (SELECT @publication:= -1) p,
  (SELECT @community:= -1) c,
  (SELECT 
        publication.name as publication_name,
        publication.community_id as community_id,
        feed_item.publication_id as publication_id,
        IF(publication.scheduled_at is not null, publication.scheduled_at, publication.created_at) as effective_publishing_date
   FROM feed_items feed_item
   JOIN publications publication ON feed_item.publication_id = publication.publication_id
   WHERE feed_item.user_id = 489387
   ORDER BY publication.community_id, effective_publishing_date ASC
  ) data_table;