如何为另一个查询的每一行使用变量运行查询

Question

我有一个桌子工作室，每个工作室可以有一个parentStudio。

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我需要查询符合某些条件的工作室，但是我还需要让所有找到的工作室的所有父母都拥有。我查询了符合条件的工作室，也查询了所有已知工作室的父母。我不知道如何将它们组合成一个查询。

我用来获取所有符合条件的工作室的查询基本上是这样的：

+----+------+--------------+----------+
| id | name | parentStudio | criteria |
+----+------+--------------+----------+
| 0  | A    | 6            | 0        |
+----+------+--------------+----------+
| 1  | B    | 0            | 1        |
+----+------+--------------+----------+
| 2  | C    | null         | 0        |
+----+------+--------------+----------+
| 3  | D    | 2            | 1        |
+----+------+--------------+----------+
| 4  | E    | null         | 1        |
+----+------+--------------+----------+
| 5  | F    | null         | 0        |
+----+------+--------------+----------+
| 6  | G    | null         | 0        |
+----+------+--------------+----------+

我用来获取所有父母的查询是：

SELECT id FROM Studios WHERE criteria = 1
+----+
| id |
+----+
| 1  |
+----+
| 3  |
+----+
| 4  |
+----+

我要寻找的结果看起来很天真：

SELECT
    @currentId AS _id,
    (
        SELECT
            name
        FROM Studios
        WHERE id = _id
    ) AS name,
    (
        SELECT
            @currentId := parentStudio
        FROM Studios
        WHERE id = _id
    ) AS parentStudio,
    @level := @level + 1 AS level
FROM
    (
        SELECT
            @currentId := 1, --starting point
            @level := 0
    ) AS vars,
    Studios AS s
WHERE
    @currentId IS NOT NULL
+----+------+--------------+-------+
| id | name | parentStudio | level |
+----+------+--------------+-------+
| 1  | B    | 0            | 0     |
+----+------+--------------+-------+
| 0  | A    | 6            | 1     |
+----+------+--------------+-------+
| 6  | G    | null         | 2     |
+----+------+--------------+-------+

但是显然，这是行不通的，因为SELECT DISTINCT _id AS id, name, parentStudio, level FROM (SELECT @currentId AS _id, ( SELECT name FROM Studios WHERE id = _id ) AS name, ( SELECT @currentId := parentStudio FROM Studios WHERE id = _id ) AS parentStudio, @level := @level + 1 AS level FROM ( SELECT @currentId := (SELECT id FROM Studios WHERE criteria = 1), @level := 0 ) AS vars, Studios AS s WHERE @currentId IS NOT NULL ) 提供了多个结果。

所需的结果集如下所示：

(SELECT id FROM Studios WHERE criteria = 1)

请注意，结果集中没有ID 5（名称F）。 它本身不符合条件，也不是符合条件的任何工作室的祖先。

有什么方法可以仅在SQL（MySQL 5）中完成吗？