如何保護孩子在父母后面出現的順序

預期順序：55,57,58,59,60,56 - 因此整個第一條父母答覆及其所有子女在第二條父母答覆前出現下面的SQL查詢返回的結果如何保護孩子在父母后面出現的順序

WITH RECURSIVE t(replyid, replypid, depth, path, reply, replied, reply_userid) AS (
    (SELECT replyid, replypid, 0, array[replyid], reply, replied, replies.userid, u.displayname, u.email_address, 
     (SELECT COUNT(*) FROM reply_revs WHERE replyid = replies.replyid) AS reply_revs 
     FROM replies 
     LEFT OUTER JOIN users u ON (replies.userid = u.userid) 
     WHERE replypid is NULL AND postid = 31 ORDER BY replied) 
    UNION ALL 
    (SELECT r.replyid, r.replypid, t.depth+1, t.path || r.replypid, r.reply, r.replied, r.userid, u.displayname, u.email_address, 
     (SELECT COUNT(*) FROM reply_revs WHERE replyid = r.replyid) 
     FROM replies r 
     JOIN t ON (r.replypid = t.replyid) 
     LEFT OUTER JOIN users u ON (r.userid = u.userid) 
     ORDER BY replied) 
) SELECT * FROM t 

replyid  replypid depth path   reply        replied 
55  NULL  0  {55}  1st parent reply     2011-02-13 11:40:48.072148-05 
56  NULL  0  {56}  2nd parent reply     2011-02-13 11:41:00.610033-05 
57  55   1  {55,55}  1st child to 1st parent reply   2011-02-13 11:41:26.541024-05 
58  55   1  {55,55}  2nd child to 1st parent reply   2011-02-13 11:41:39.485405-05 
59  55   1  {55,55}  3rd child to 1st parent reply   2011-02-13 11:41:51.35482-05 
60  59   2  {55,55,59} 1st child to 3rd child of 1st parent reply 2011-02-13 11:42:14.866852-05

的錯誤的順序然而，僅僅在套結「ORDER BY路徑」到最後修復了這個，但僅限於按升序排列

WITH RECURSIVE t(replyid, replypid, depth, path, reply, replied, reply_userid) AS (
    (SELECT replyid, replypid, 0, array[replyid], reply, replied, replies.userid, u.displayname, u.email_address, 
     (SELECT COUNT(*) FROM reply_revs WHERE replyid = replies.replyid) AS reply_revs 
     FROM replies 
     LEFT OUTER JOIN users u ON (replies.userid = u.userid) 
     WHERE replypid is NULL AND postid = 31 ORDER BY replied) 
    UNION ALL 
    (SELECT r.replyid, r.replypid, t.depth+1, t.path || r.replypid, r.reply, r.replied, r.userid, u.displayname, u.email_address, 
     (SELECT COUNT(*) FROM reply_revs WHERE replyid = r.replyid) 
     FROM replies r 
     JOIN t ON (r.replypid = t.replyid) 
     LEFT OUTER JOIN users u ON (r.userid = u.userid) 
     ORDER BY replied) 
) SELECT * FROM t ORDER BY path 

replyid replypid depth path   reply        replied 
55  NULL 0  {55}  1st parent reply     2011-02-13 11:40:48.072148-05 
57  55  1  {55,55}  1st child to 1st parent reply   2011-02-13 11:41:26.541024-05 
58  55  1  {55,55}  2nd child to 1st parent reply   2011-02-13 11:41:39.485405-05 
59  55  1  {55,55}  3rd child to 1st parent reply   2011-02-13 11:41:51.35482-05 
60  59  2  {55,55,59} 1st child to 3rd child of 1st parent reply 2011-02-13 11:42:14.866852-05 
56  NULL 0  {56}  2nd parent reply     2011-02-13 11:41:00.610033-05

所以讓我們試試現在DESCENDING通過附加「ORDER BY路徑DESC」結果是：

replyid replypid depth path   reply        replied 
56  NULL 0  {56}  2nd parent reply     2011-02-13 11:41:00.610033-05 
60  59  2  {55,55,59} 1st child to 3rd child of 1st parent reply 2011-02-13 11:42:14.866852-05 
57  55  1  {55,55}  1st child to 1st parent reply   2011-02-13 11:41:26.541024-05 
58  55  1  {55,55}  2nd child to 1st parent reply   2011-02-13 11:41:39.485405-05 
59  55  1  {55,55}  3rd child to 1st parent reply   2011-02-13 11:41:51.35482-05 
55  NULL 0  {55}  1st parent reply     2011-02-13 11:40:48.072148-05

現在看起來好像第一個父母答覆的孩子是第二個父母答覆的孩子。

我的問題是：如何訂購結果，以便兒童或深度大於0的結果始終出現在對應的父母身後，而不是其他父母項之後？

我想看到的結果：

replyid replypid depth path   reply        replied 
56  NULL 0  {56}  2nd parent reply     2011-02-13 11:41:00.610033-05 
55  NULL 0  {55}  1st parent reply     2011-02-13 11:40:48.072148-05 
57  55  1  {55,55}  1st child to 1st parent reply   2011-02-13 11:41:26.541024-05 
58  55  1  {55,55}  2nd child to 1st parent reply   2011-02-13 11:41:39.485405-05 
59  55  1  {55,55}  3rd child to 1st parent reply   2011-02-13 11:41:51.35482-05 
60  59  2  {55,55,59} 1st child to 3rd child of 1st parent reply 2011-02-13 11:42:14.866852-05

由於RhodiumToad在#postgresql Freenode上我能想出下面的PHP和SQL查詢其工作得！

if (isset($_SESSION["userid"])) { 
    $s_col1 = ", (SELECT COUNT(*) FROM votes WHERE replyid = replies.replyid AND userid = %d) AS reply_voted"; 
    $s_col2 = ", (SELECT COUNT(*) FROM votes WHERE replyid = r.replyid AND userid = %d)"; 
} else { $s_col1 = ""; $s_col2 = ""; } 

if ($sort == "newest") { $s_arr1 = "-extract(epoch from replied)::integer"; $s_arr2 = " || -extract(epoch from r.replied)::integer"; } 
else if ($sort == "oldest") { $s_arr1 = "extract(epoch from replied)::integer"; $s_arr2 = " || extract(epoch from r.replied)::integer"; } 
else if ($sort == "topvotes") { $s_arr1 = "-votes"; $s_arr2 = " || -r.votes"; } 
else { $s_arr1 = ""; $s_arr2 = ""; } 

$sql = "WITH RECURSIVE t(replyid, replypid, depth, path, reply, replied, reply_userid) AS (
     (SELECT replyid, replypid, 0, array[$s_arr1,replyid], reply, replied, replies.userid, u.displayname, u.email_address, 
      (SELECT COUNT(*) FROM reply_revs WHERE replyid = replies.replyid) AS reply_revs, 
      (SELECT COUNT(*) FROM votes WHERE replyid = replies.replyid) AS reply_votes 
      $s_col1 
     FROM replies 
     LEFT OUTER JOIN users u ON (replies.userid = u.userid) 
     WHERE replypid is NULL AND postid = %d) 
     UNION ALL 
     (SELECT r.replyid, r.replypid, t.depth+1, t.path$s_arr2 || r.replyid, r.reply, r.replied, r.userid, u.displayname, u.email_address, 
      (SELECT COUNT(*) FROM reply_revs WHERE replyid = r.replyid) AS reply_revs, 
      (SELECT COUNT(*) FROM votes WHERE replyid = r.replyid) AS reply_votes 
      $s_col2 
     FROM replies r 
     JOIN t ON (r.replypid = t.replyid) 
     LEFT OUTER JOIN users u ON (r.userid = u.userid)) 
    ) SELECT * FROM t ORDER BY path";

來源

2011-02-14 Miz

什麼讓孩子成爲第一個孩子？如果是回覆的日期，則必須按此值排序。

來源

2011-02-14 19:36:58

在第一組結果中，第一個孩子（或深度爲1的第一行）是第一個按照您的建議回答的日期，但是，您建議即使它已經是我必須按這個值來訂購？第二個內部SELECT語句具有「ORDER BY已回答」子句。否則，如果您的意思是添加到最後一行「ORDER BY路徑，回覆DESC」，那麼第一個答覆仍然首先出現。 – Miz 2011-02-14 21:53:07

在最後一次查詢中，您確實有兩種排序。父母可以升序或降序排序，但孩子只能升序排序。

看過這個後，我相信你可以得到一個像這樣的解決方案。

order by case 
     when depth = 0 
      then path 
    /* 
     secret function that always returns the 
     right numbers regardless of whether or not the sort is ascending. 
    */ 
     else XXX_function('DESC', path) 
    end desc;

我認爲邏輯是健全的，但你必須弄清楚如何更換降序排序，因爲事情會「倒掛」的數字。（可能會顛倒陣列位置）

來源

2011-02-14 23:02:18

如何保護孩子在父母后面出現的順序

回答

相關問題