PSQL dense_rank中不同項目中的一個項目

我可以使用以下PSQL查詢（最後）獲取我表格中單行的dense_rank，但是，我希望能夠顯示此爲：PSQL dense_rank中不同項目中的一個項目

dense_rank OUT OF total distinct ranks

例如，由於dense_rank允許「關係」，可以這麼說，如果我有100行和所選行排名第14位（也有隻有59不同等級），我想說：

Ranked 14th out of 59

有沒有一種方法可以修改我的查詢來實現這一目標，還是必須使用多個查詢？

這裏是我的查詢：

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 = ?

來源

2017-05-30 lightyrs

你可以試試這個。

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=?

來源

2017-05-30 21:16:37

這似乎工作@vkp，謝謝！如果你有時間，你能解釋這個黑魔法嗎？非常感激。 – lightyrs

@lightyrs ..我添加的唯一附加功能是獲得最高排名（即本例中爲59）的MAX（RNK）OVER（）。 'RNK'就是你已經計算好的。連接它們會得到你所需要的。 –

基本上，你想這樣：

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，因爲Postgres的不支持COUNT(DISTINCT)作爲窗口函數。您可以使用其他窗口功能實現它：\

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 = ?;

來源

2017-05-30 21:23:36

PSQL dense_rank中不同項目中的一個項目

回答

相關問題