MySQL，获取一行在多行集合中的位置

Question

我有这个查询计算 publication 在一组出版物中的位置（这里命名为 community），根据 effective_publishing_date :

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;

此结果为：

[![在此处输入图片描述][1]][1]

现在我有一个 feed_item 的列表，其中每个 community_id 和 publication_id 作为属性，我想为每个 feed_item 获取关联的 publication_rank.

例如，如果我有一个 publication_item，其中 publication_id = 18 且 community_id = 2，我想要 publication_id 的 publication_rank #在 community_id #2 的所有出版物中排名第 18。我没有在一个查询（或子查询等）中成功获得它。

提前致谢，

Answer 1

这是我最终找到的两个解决方案。感谢@Barmar！

仅限连接：

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;

有 SQL 个变量

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;

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