Question

通过回复的预期顺序：55,57,58,59,60,56 - 这样整个第一个家长回复及其所有子女出现在第二个家长回复之前以下SQL查询返回错误的结果顺序

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

然而，仅仅在“ORDER BY路径”上添加到最后修复此问题，但仅限于ASCENDING命令

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

所以让我们现在尝试DESCENDING而不是附加“ORDER BY path DESC”结果是：

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

现在好像第一个家长回复的孩子是第二个家长回复的孩子。

我的问题是：我如何订购结果，以便儿童或结果深度＆gt; 0总是出现在相应的父母之后，而不是在其他父项之后？

我想看到的结果：

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

感谢Freenode上的#postgresql中的RhodiumToad，我能够提出以下PHP和SQL查询，这些查询非常适合！

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";

Answer 1

您在上一次查询中确实有两种排序。父母可以按升序或降序排序，但孩子只能按升序排序。

看了之后我相信你可以得到这样的解决方案。

   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;

我认为逻辑是合理的，但你必须弄清楚如何更换降序的数字，因为事情会“颠倒”。（也许反转阵列位置）

Answer 2

是什么让孩子成为第一个孩子？如果是回复日期，您还必须按此值排序。

如何保持孩子在父母身后出现的秩序

2 个答案: