0

replyid の予期される順序: 55、57、58、59、60、56 -- 1 番目の親の返信全体とそのすべての子が 2 番目の親の返信の前に表示されるようにします。次の SQL クエリは間違った順序の結果を返します。

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

ただし、「ORDER BYパス」を最後に追加するだけでこれが修正されますが、昇順のみです

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

それでは、代わりに「ORDER BY path DESC」を追加して、DESCENDING を試してみましょう。結果は次のとおりです。

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

これで、1 番目の親の返信に対する子が 2 番目の親の返信の子であるかのように表示されます。

私の質問は次のとおりです: 子または深さ > 0 の結果が常に対応する親の後に表示され、他の親アイテムの後に表示されるように、結果を並べ替えるにはどうすればよいですか?

私が見たい結果：

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

Freenode の #postgresql にある RhodiumToad のおかげで、驚くほど機能する次の PHP および SQL クエリを思いつくことができました!

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";
4

2 に答える 2

1

最後のクエリでは、実際には 1 つに 2 つの並べ替えがあります。親は昇順または降順でソートできますが、子は昇順でしかソートできません。

これを見た後、私はあなたがこのようなもので解決策を得ることができると信じています.

   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;

論理は正しいと思いますが、物事が「逆さま」になるため、降順ソートで数値を置き換える方法を理解する必要があります。（配列の位置を逆にするかもしれません）

于 2011-02-14T23:02:18.437 に答える
0

子供が最初の子供になる理由は何ですか? 返信日の場合は、この値でも注文する必要があります。

于 2011-02-14T19:36:58.553 に答える