Sql 递归查询："；表不包括'；“不存在”；内部功能，但外部功能

我试图在Ubuntu 18.04上的MariaDB 10.3.7上的函数中封装一个递归临时表查询。我试着把陈述分解成更基本的部分，但在我把它们放在一起之前，它们都是正确的

这是我想将语句转换为函数的一个示例：

with recursive Descendants as (
    select * from Characters where id = 91402
    union
    select c.* from Characters as c, Descendants as d
        where d.id = c.mother_id or d.id = c.real_father_id
) select count(distinct(id)) from Descendants

create function count_descendants(cid int unsigned) returns int unsigned return (
    with recursive Descendants as (
        select * from Characters where id = cid
        union
        select c.* from Characters as c, Descendants as d
            where d.id = c.mother_id or d.id = c.real_father_id
    ) select count(distinct(id)) from Descendants
);

它打印出后代字符的数量，这就是我想要的

以下是我到目前为止将其转化为函数的尝试：

with recursive Descendants as (
    select * from Characters where id = 91402
    union
    select c.* from Characters as c, Descendants as d
        where d.id = c.mother_id or d.id = c.real_father_id
) select count(distinct(id)) from Descendants

create function count_descendants(cid int unsigned) returns int unsigned return (
    with recursive Descendants as (
        select * from Characters where id = cid
        union
        select c.* from Characters as c, Descendants as d
            where d.id = c.mother_id or d.id = c.real_father_id
    ) select count(distinct(id)) from Descendants
);

Maria接受这一点，但如果我尝试使用它：

select count_descendants(91402);

它将打印此错误消息：

ERROR 1146 (42S02) at line 12: Table 'ck2.Characters' doesn't exist

“ck2”是正在使用的数据库的名称。如果有人想看到，我可以发布模式

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编辑：我已经在MariaDB的bug跟踪器上以bug的形式提交了它：

这是一个

尝试用数据库和模式名称指定表。“ck2.dbo.Characters”我不知道这是什么，但在名称中包含“dbo”会产生语法错误。使用“ck2.Characters”的结果与不包含数据库名称的结果相同。然后尝试将表名“Characters”替换为“ck2.Characters”，这就是我尝试的，没有任何区别。我不认为这是个问题，因为我已经

使用了ck2在脚本的开头，我正在运行这些。