PSQL dense_rank中不同项目中的一个项目

我可以使用以下PSQL查询（最后）获取我表格中单行的dense_rank，但是，我希望能够显示此为：PSQL dense_rank中不同项目中的一个项目

dense_rank OUT OF total distinct ranks

例如，由于dense_rank允许“关系”，可以这么说，如果我有100行和所选行排名第14位（也有只有59不同等级），我想说：

Ranked 14th out of 59

有没有一种方法可以修改我的查询来实现这一目标，还是必须使用多个查询？

这里是我的查询：

SELECT ranked.* 
FROM 
    (SELECT id, 
      postable_id, 
      spread_count, 
      bury_count, 
      read_count, 
      (spread_count*3) + (bury_count*-2) + (read_count*-1) AS score, 
      dense_rank() OVER (
          ORDER BY (spread_count*3) + (bury_count*-2) + (read_count*-1) DESC) AS RANK 
    FROM posts) AS ranked 
WHERE id = ?

来源

2017-05-30 lightyrs

你可以试试这个。

SELECT t.* 
FROM (SELECT ranked.*, 
     RNK||' out of '||MAX(RNK) OVER() as rnk_pos 
     FROM 
     (SELECT id, 
      postable_id, 
      spread_count, 
      bury_count, 
      read_count, 
      (spread_count*3) + (bury_count*-2) + (read_count*-1) AS score, 
      dense_rank() OVER (
          ORDER BY (spread_count*3) + (bury_count*-2) + (read_count*-1) DESC) AS RNK 
     FROM posts) AS ranked 
    ) t 
WHERE id=?

来源

2017-05-30 21:16:37

这似乎工作@vkp，谢谢！如果你有时间，你能解释这个黑魔法吗？非常感激。 – lightyrs

@lightyrs ..我添加的唯一附加功能是获得最高排名（即本例中为59）的MAX（RNK）OVER（）。 'RNK'就是你已经计算好的。连接它们会得到你所需要的。 –

基本上，你想这样：

SELECT p.* 
FROM (SELECT p.* 
      (spread_count*3) + (bury_count*-2) + (read_count*-1) AS score, 
      dense_rank() OVER (ORDER BY (spread_count*3) + (bury_count*-2) + (read_count*-1) DESC) AS RANK, 
      count(distinct (spread_count*3) + (bury_count*-2) + (read_count*-1)) over() as outof 
     FROM posts p 
    ) p 
WHERE id = ?;

唉，不行，因为Postgres的不支持COUNT(DISTINCT)作为窗口函数。您可以使用其他窗口功能实现它：\

SELECT p.* 
FROM (SELECT p.*, 
      dense_rank() over (order by score desc) AS RANK, 
      sum((seqnum = 1)::int) as outof 
     FROM (SELECT p.*, 
        (spread_count*3) + (bury_count*-2) + (read_count*-1) AS score, 
        row_number() over (partition by (spread_count*3) + (bury_count*-2) + (read_count*-1) AS score order by spread_scount) as seqnum 
      FROM posts p 
      ) p 
    ) p 
WHERE id = ?;

来源

2017-05-30 21:23:36

PSQL dense_rank中不同项目中的一个项目

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