Данные цепочки запросов в MySQL

Используйте рекурсивный CTE:

WITH RECURSIVE
cte1 AS (
  SELECT col1, col2, CAST(col2 AS CHAR(65535)) path, 1 level
  FROM test
  UNION ALL
  SELECT cte1.col1, cte1.col2, CONCAT_WS(',', cte1.path, test.col2), cte1.level + 1
  FROM test
  JOIN cte1 ON FIND_IN_SET(test.col1, cte1.path)
  WHERE NOT FIND_IN_SET(test.col2, cte1.path)
  ),
cte2 AS (
  SELECT *, ROW_NUMBER() OVER (PARTITION BY col1, col2 ORDER BY LENGTH(path) DESC) rn
  FROM cte1
  )
SELECT col1, col2, SUBSTRING_INDEX(path, ',', -1) last_chain_data
FROM cte2 
WHERE rn = 1
ORDER BY last_chain_data, level DESC;

рабочий пример

Еще один пример рекурсивного запроса

with RECURSIVE trn as(select *,row_number()over(order by col1,col2) rn from test)
, r as(
  select 0 lvl,t1.rn,t1.col1,t1.col2,t2.col2 next, t2.rn next_rn
    ,concat(':',t1.rn,':',t2.rn) path
  from trn t1 left join trn t2
     on t2.col1=t1.col2 and t1.rn<>t2.rn
  union all
  select lvl+1 lvl,r.rn,r.col1,r.col2,t.col2 next,t.rn 
    ,concat(path,':',t.rn) path
  from r left  join trn t
     on t.col1=r.next and t.rn<>r.rn
  where lvl<7 and locate(t.rn,path)=0
)
select *
  from(
  select * 
     ,row_number()over(partition by rn order by lvl desc)rrn
  from r
  ) x
where rrn=1;

Выход:

О (765, 765) --> null - почему результат должен быть таким, должно быть описано далее в вопросе. Я не хочу гадать. В этом случае результат может быть неоднозначным.

рабочий пример

Столбец действительно мог бы выиграть от столбца id. Но можно и без этого:

create table Table_A (col1 int, col2 int);
insert Table_A values
(881,113),
(988,899),
(113,765),
(765,765),
(122,881),
(300,400),
(765,910),
(910,345),
(999,988);

Вот запрос:

with recursive
i as    (select row_number() over() as id,col1,col2
        from table_a), -- as we don't have an id column, we add one here. The table i will be used later on instead of the orginal table
c as    (
        select t1.id, t1.col1,t1.col2
        from i t1 
        left join i t2
        on t1.col2=t2.col1
        where t1.col1<>t1.col2 and t2.col1 is not null
        union 
        select c.id, t.col1,t.col2
        from i t
        right join c
        on c.col2=t.col1
        where t.col1 is not null and t.col1<>t.col2 -- if t.col1<>t.col2 is removed here, the recursive process will have meaningless calculation when t.col1<>t.col2    
) -- table c is to calculate for those rows which have a chained row, also the rows whose col1<>col2
select col1,col2,last_chain_data -- this outter layer is used to make sure each row appears in their original order from the base table, as UNION statement doesn't enforce the feature of ORDER BY clause within it) 
from (select distinct id,first_value(col1)  over w as col1,first_value(col2)  over w as col2,last_value(col2) over w as last_chain_data
     from c  window w as (partition by id) -- the first query in the UNION is to get the required output for those rows which have a chained row, also the rows whose col1<>col2
     UNION
     select distinct t1.id,t1.col1,t1.col2,null
     from i t1 
     left join i t2
     on t1.col2=t2.col1  
     where t1.col1=t1.col2 or t2.col1 is null  -- contrary to the 1st query, the second query in the UNION is for those rows which don't have a chained row, also the rows whose col1=col2
) tb
order by id
;

-- result
+------+------+-----------------+
| col1 | col2 | last_chain_data |
+------+------+-----------------+
|  881 |  113 |             345 |
|  988 |  899 |            NULL |
|  113 |  765 |             345 |
|  765 |  765 |            NULL |
|  122 |  881 |             345 |
|  300 |  400 |            NULL |
|  765 |  910 |             345 |
|  910 |  345 |            NULL |
|  999 |  988 |             899 |
+------+------+-----------------+

См. https://www.db-fiddle.com/f/o1sJwMD7DFRg7iQc2wnJsU/1