MYSQL 8 - 如何在检索树时在 CTE 中找出“节点”是否为“叶节点”

如何解决MYSQL 8 - 如何在检索树时在 CTE 中找出“节点”是否为“叶节点”

我有以下数据：

MysqL> SELECT id,parent_id,name FROM account_type WHERE account_classification_id = 1 LIMIT 7;
+----+-----------+----------------------+
| id | parent_id | name                 |
+----+-----------+----------------------+
| 30 |      NULL | AKTIVA               |
| 40 |        30 | Aktiva Lancar        |
| 41 |        40 | Kas & Bank           |
| 42 |        41 | Kas                  |
| 43 |        41 | Kas Tunai USD        |
| 44 |        41 | Kas Tunai Di Brankas |
| 46 |        41 | Bank                 |
+----+-----------+----------------------+
7 rows in set (0.00 sec)

因此，使用 CTE：

WITH RECURSIVE account_path (root,id,name,lvl,`sort`,account_roll_up_id) AS
                   (
                       SELECT
                              id AS root,`name`,0 lvl,account_roll_up_id

                       FROM account_type
                                WHERE id IN (30,213)
                       UNION ALL

                       SELECT
                              ap.root,at.id,at.parent_id,at.name,(ap.lvl + 1),at.sort,at.account_roll_up_id

                       FROM account_path AS ap
                                JOIN account_type AS at ON ap.id = at.parent_id
                   )

SELECT * FROM `account_path`
ORDER BY account_path.sort
;

+------+------+-----------+----------------------+------+------+--------------------+
| root | id   | parent_id | name                 | lvl  | sort | account_roll_up_id |
+------+------+-----------+----------------------+------+------+--------------------+
|   30 |   30 |      NULL | AKTIVA               |    0 |    1 |               NULL |
|   30 |   40 |        30 | Aktiva Lancar        |    1 |    2 |                  2 |
|   30 |   41 |        40 | Kas & Bank           |    2 |    3 |                  2 |
|   30 |   42 |        41 | Kas                  |    3 |    4 |                  2 |
|   30 |   43 |        41 | Kas Tunai USD        |    3 |    5 |                  2 |
|   30 |   44 |        41 | Kas Tunai Di Brankas |    3 |    6 |                  2 |
|   30 |   46 |        41 | Bank                 |    3 |    7 |                  2 |
+------+------+-----------+----------------------+------+------+--------------------+
7 rows in set (0.01 sec)

当我检索所有节点时，我需要每一行都有一个布尔值来确定 is_a_leaf_node， 是否有可能？ 像这样：

+------------+------+------+-----------+----------------------+------+------+--------------------+
|  is_a_leaf | root | id   | parent_id  | name                 | lvl  | sort| account_roll_up_id |
+------------+------+------+------------+----------------------+------+-----+--------------------+
|       FALSE|   30 |   30 |      NULL | AKTIVA               |    0 |    1 |               NULL |
|       FALSE|   30 |   40 |        30 | Aktiva Lancar        |    1 |    2 |                  2 |
|       FALSE|   30 |   41 |        40 | Kas & Bank           |    2 |    3 |                  2 |
|       TRUE |   30 |   42 |        41 | Kas                  |    3 |    4 |                  2 |
|       TRUE |   30 |   43 |        41 | Kas Tunai USD        |    3 |    5 |                  2 |
|       TRUE |   30 |   44 |        41 | Kas Tunai Di Brankas |    3 |    6 |                  2 |
|       TRUE |   30 |   46 |        41 | Bank                 |    3 |    7 |                  2 |
+------------+------+------+-----------+----------------------+------+------+--------------------+

解决方法

已解决，谢谢@Akina

WITH RECURSIVE account_path (is_a_leaf,root,id,parent_id,name,lvl,`sort`,account_roll_up_id) AS
                   (
                       SELECT id    is_a_leaf,id AS root,`name`,0     lvl,account_roll_up_id

                       FROM account_type
                       WHERE id IN (30,213)
                       UNION ALL

                       SELECT (SELECT EXISTS(
                                              SELECT parent_id
                                              FROM account_type
                                              WHERE parent_id = at.id
                                          )),ap.root,at.id,at.parent_id,at.name,(ap.lvl + 1),at.sort,at.account_roll_up_id

                       FROM account_path AS ap
                                JOIN account_type AS at ON ap.id = at.parent_id
                   )

SELECT *
FROM `account_path`
ORDER BY account_path.sort
LIMIT 7;