我正在尝试通过CASE语句进行计算，这些语句依赖于上一行的计算结果。我正在使用的数据是分层数据。我的最终目标是使结果数据的结构与Modified Preorder Tree Traversal algorithm相符。
这是我的原始数据：
+-------+--------+
| id | parent |
+-------+--------+
| 1 | (null) |
+-------+--------+
| 600 | 1 |
+-------+--------+
| 690 | 600 |
+-------+--------+
| 6990 | 690 |
+-------+--------+
| 6900 | 690 |
+-------+--------+
| 69300 | 6900 |
+-------+--------+
| 69400 | 6900 |
+-------+--------+
这就是我想要的最终结果。我很高兴继续解释为什么这是我要寻找的与MPTT等相关的内容。
+-------+-----------+-----+------+--+--+--+--+
| id | parent_id | lft | rght | | | | |
+-------+-----------+-----+------+--+--+--+--+
| 1 | | 1 | 14 | | | | |
+-------+-----------+-----+------+--+--+--+--+
| 600 | 1 | 2 | 13 | | | | |
+-------+-----------+-----+------+--+--+--+--+
| 690 | 600 | 3 | 12 | | | | |
+-------+-----------+-----+------+--+--+--+--+
| 6900 | 690 | 4 | 9 | | | | |
+-------+-----------+-----+------+--+--+--+--+
| 6990 | 690 | 10 | 11 | | | | |
+-------+-----------+-----+------+--+--+--+--+
| 69300 | 6900 | 5 | 6 | | | | |
+-------+-----------+-----+------+--+--+--+--+
| 69400 | 6900 | 7 | 8 | | | | |
+-------+-----------+-----+------+--+--+--+--+
这是到目前为止我的SQL代码。它计算了我认为下面描述的算法所需的许多字段。这是企业设置内的“组织”数据，这就是为什么orgn缩写在我的代码中很常见的原因。
以下是我认为可以成功将其转换为MPTT格式的算法：
-If level is root (lvl=1), lft = 1, rght = subnodes*2 + 2
-If level is the next level down (lvl = prev_lvl+1), and prev_parent != parent (meaning this is the first sibling)
-lft = parent_lft+1
-If lvl = prev_lvl, so we are on the same level (don’t know if this is a true sibling of the same parent yet)
-if parent = prev_parent, lft=prev_rght+1 (true sibling, just use previous sibling’s right + 1)
-if parent != prev_parent, lft=parent_lft+1 (same level, not true sibling, so use parent’s left + 1)
-rght=(subnodes*2) + lft + 1
SQL代码：
WITH tab1 (
id,
parent_id
) AS (
SELECT
1,
NULL
FROM
dual
UNION ALL
SELECT
600,
1
FROM
dual
UNION ALL
SELECT
690,
600
FROM
dual
UNION ALL
SELECT
6990,
690
FROM
dual
UNION ALL
SELECT
6900,
690
FROM
dual
UNION ALL
SELECT
69300,
6900
FROM
dual
UNION ALL
SELECT
69400,
6900
FROM
dual
),t1 (
id,
parent_id,
lvl
) AS (
SELECT
id,
parent_id,
1 AS lvl
FROM
tab1
WHERE
parent_id IS NULL
UNION ALL
SELECT
t2.id,
t2.parent_id,
lvl + 1
FROM
tab1 t2,
t1
WHERE
t2.parent_id = t1.id
)
SEARCH BREADTH FIRST BY id SET order1,orgn_subnodes AS (
SELECT
id AS id,
COUNT(*) - 1 AS subnodes
FROM
(
SELECT
CONNECT_BY_ROOT ( t1.id ) AS id
FROM
t1
CONNECT BY
PRIOR t1.id = t1.parent_id
)
GROUP BY
id
),orgn_partial_data AS (
SELECT
orgn_subnodes.id AS id,
orgn_subnodes.subnodes,
parent_id,
lvl,
LAG(lvl,1) OVER(
ORDER BY
order1
) AS prev_lvl,
LAG(parent_id,1) OVER(
ORDER BY
order1
) AS prev_parent,
CASE
WHEN parent_id IS NULL THEN 1
END
lft,
CASE
WHEN parent_id IS NULL THEN ( subnodes * 2 ) + 2
END
rght,
order1
FROM
orgn_subnodes
JOIN t1 ON orgn_subnodes.id = t1.id
) SELECT
*
FROM
orgn_partial_data;
结果是：
+-------+----------+-----------+-----+----------+-------------+-----+------+--------+
| id | subnodes | parent_id | lvl | prev_lvl | prev_parent | lft | rght | order1 |
+-------+----------+-----------+-----+----------+-------------+-----+------+--------+
| 1 | 6 | | 1 | | | 1 | 14 | 1 |
+-------+----------+-----------+-----+----------+-------------+-----+------+--------+
| 600 | 5 | 1 | 2 | 1 | | | | 2 |
+-------+----------+-----------+-----+----------+-------------+-----+------+--------+
| 690 | 4 | 600 | 3 | 2 | 1 | | | 3 |
+-------+----------+-----------+-----+----------+-------------+-----+------+--------+
| 6900 | 2 | 690 | 4 | 3 | 600 | | | 4 |
+-------+----------+-----------+-----+----------+-------------+-----+------+--------+
| 6990 | 0 | 690 | 4 | 4 | 690 | | | 5 |
+-------+----------+-----------+-----+----------+-------------+-----+------+--------+
| 69300 | 0 | 6900 | 5 | 4 | 690 | | | 6 |
+-------+----------+-----------+-----+----------+-------------+-----+------+--------+
| 69400 | 0 | 6900 | 5 | 5 | 6900 | | | 7 |
+-------+----------+-----------+-----+----------+-------------+-----+------+--------+
我不在乎树中“兄弟节点”的顺序。另外，如果您发现我开始使用的SQL没有用，则可以发布不使用其中任何一个的答案。我只发布了我认为我需要执行算法步骤的信息。
我将接受任何Oracle代码（数据库过程，SELECT语句等）作为答案。
如果需要它们，请询问更多细节！
答案 0 :(得分：1)
我认为帖子开头有错别字，对于69400应该是（7，8），而不是（4，8）。
获得结果的规范方法是使用递归过程/函数。 下面的方法使用过程和临时表，但是您可以使用返回集合的函数来实现相同的目的。
临时表
create global temporary table tmp$ (id int, l int, r int) on commit delete rows;
包装
create or replace package pkg as
procedure p(p_id in int);
end pkg;
/
sho err
包装体
create or replace package body pkg as
seq int;
procedure p_(p_id in int) as
begin
seq := seq + 1;
insert into tmp$(id, l, r) values (p_id, seq, null);
for i in (select id from tab1 where parent_id = p_id order by id) loop
p_(i.id);
end loop;
seq := seq + 1;
update tmp$ set r = seq where id = p_id;
end;
procedure p(p_id in int) as
begin
seq := 0;
p_(p_id);
end;
end pkg;
/
sho err
在SQL * PLus中测试
SQL> exec pkg.p(1);
PL/SQL procedure successfully completed.
SQL> select * from tmp$;
ID L R
---------- ---------- ----------
1 1 14
600 2 13
690 3 12
6900 4 9
69300 5 6
69400 7 8
6990 10 11
7 rows selected.
更新
没有全局变量的独立过程
create or replace procedure p(p_id in int, seq in out int) as
begin
seq := seq + 1;
insert into tmp$(id, l, r) values (p_id, seq, null);
for i in (select id from tab1 where parent_id = p_id order by id) loop
p(i.id, seq);
end loop;
seq := seq + 1;
update tmp$ set r = seq where id = p_id;
end;
/
在SQL * PLus中测试
SQL> var n number
SQL> exec :n := 0;
PL/SQL procedure successfully completed.
SQL> exec p(1, :n);
PL/SQL procedure successfully completed.
SQL> select * from tmp$;
ID L R
---------- ---------- ----------
1 1 14
600 2 13
690 3 12
6900 4 9
69300 5 6
69400 7 8
6990 10 11
7 rows selected.