这是一个典型的图形遍历问题-需要一个递归查询，从MariaDB 10.2.2开始可用。

这是一种方法，首先将边缘构建为连接节点的元组，然后迭代遍历数据集，并跟踪已访问的节点。我们可以通过跟踪最小访问节点来确定每个节点属于哪个组。

with recursive 
    edges as (
        select t1.lagr_number as lagr_number1, t2.lagr_number as lagr_number2
        from mytable t1
        inner join mytable t2 on t2.ltran_number = t1.ltran_number
    ),
    cte as (
        select lagr_number1, lagr_number2, concat(lagr_number1, ',', lagr_number2) as visited
        from edges
        union all
        select c.lagr_number1, e.lagr_number2, concat(c.visited, ',', e.lagr_number2)
        from cte c
        inner join edges e on e.lagr_number1 = c.lagr_number2
        where not find_in_set(e.lagr_number2, c.visited)
    )
select lagr_number1 as lagr_number, dense_rank() over(order by min(lagr_number2)) as grp
from cte
group by lagr_number1
order by grp, lagr_number1

DB Fiddle上的演示

样本数据：

lagr_number | ltran_number
：---------- | ：-----------
K000001 | V000001     
K000001 | V000004     
K000001 | V000005     
K000002 | V000001     
K000003 | V000002     
K000003 | V000003     
K000004 | V000005     
K000005 | V000007     
K000005 | V000008     
K000006 | V000009     
K000007 | V000009     

结果：

lagr_number | grp
：---------- | -：
K000001 | 1个
K000002 | 1个
K000004 | 1个
K000003 | 2个
K000005 | 3
K000006 | 4
K000007 | 4