MySQL，获取多行集合中一行的位置

我有一个查询，它计算了一个出版物在一组出版物中的位置，这些出版物在这里被命名为community，根据有效发布日期：

结果如下：

[！[在此处输入图像描述][1][1]

现在我有了一个feed_项列表，其中每个community_id和publication_id都是属性，我想为每个feed_项获取相关的publication_排名

例如，如果我有一个publication_项，publication_id=18，community_id=2，那么我需要publication_id 18在community_id 2的所有发布中的publication_排名。我不能在一个查询或子查询中成功地得到它

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谢谢大家，这里是我最终找到的两个解决方案。感谢@Barmar

仅适用于连接： 使用SQL变量

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以防万一。。。您使用的是哪个版本的MySQL？您好@GMB，我有MySQL 5.7.26，请参阅如何在组内获得排名。然后你可以加入你的feed_项目列表，以获得其在组中的排名。谢谢你@Barmar，我解决了我的问题，谢谢你的链接：-很好的研究和适应工作！

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;

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;

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;