Recursive MySQL Query

MYSQL Recursive query with conditions

This query gets you the price hirarchically:

with recursive cte(id_product, rate, price, origin_rate) as
(
  select id_product, rate, price, rate
  from mytable
  union all
  select cte.id_product, cte.rate, t.price, t.rate
  from cte
  join map on map.origin_rate = cte.origin_rate
  left join mytable t on t.id_product = cte.id_product
                     and t.rate = map.destination_rate
  where cte.price is null
)
select id_product, rate, price
from cte
where price is not null
order by id_product, rate;

How to use recursion in MYSQL View

You may use a recursive CTE inside a view:

CREATE VIEW categories_view AS
WITH RECURSIVE cte AS (
    SELECT id, parentId, name, name AS path
    FROM categories
    WHERE parentId IS NULL
    UNION ALL
    SELECT c.id, c.parentId, c.name, CONCAT_WS(' > ', t.path, c.name)
    FROM categories c
    INNER JOIN cte t ON c.parentId = t.id
)

SELECT *
FROM cte
ORDER BY id;

Here is a demo showing that the logic is working.

MySql Recursive - get all children and parents from a given id

I would use two separate recursive queries: one to bring the children, the other for the parents, and then union the results. You can keep track of the level of each node to properly order the records int he resultset:

with recursive 
    children as (
        select 1 as lvl, d.* from department d where id = 6
        union all
        select c.lvl, d.* from department d inner join children c on c.id = d.father
    ),
    parents as (
        select 1 as lvl, d.* from department d where id = 6
        union all
        select p.lvl - 1, d.* from department d inner join parents p on d.id = p.father
    )
select * from parents
union   -- on purpose, to remove the duplicate on id 6
select * from children
order by lvl;

This is safer than having multiple union all members in the same query. MySQL does not guarantee the order of evaluation of the members in the recursion, so using this technique could lead to unexpected behavior.

Demo on DB Fiddle

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Unrelated to your question, but: the following can be seen in your code:

SET GLOBAL sql_mode=(SELECT REPLACE(@@sql_mode,'ONLY_FULL_GROUP_BY',''));
SET SESSION sql_mode = '';

Just don't. ONLY_FULL_GROUP_BY is there for a good reason, that is to have MySQL behave consistenly with the SQL standard as regard to aggregation query. Disabling this SQL mode is never a good idea.

MySQL - Recursively list all parents and ancestors of all items in table

If you are running MySQL 8.0, this is best solved with a recursive query:

with recursive cte as (
    select id, parent_id, 1 lvl from mytable
    union all
    select c.id, t.parent_id, lvl + 1
    from cte c
    inner join mytable t on t.id = c.parent_id
)
select id, group_concat(parent_id order by lvl) all_parents
from cte
group by id

Demo on DB Fiddle:

id | all_parents
-: | :----------
 1 | 0          
 2 | 0          
 3 | 0          
17 | 3,0        
31 | 17,3,0     

How to do the Recursive SELECT query in MySQL?

Edit

Solution mentioned by @leftclickben is also effective.
We can also use a stored procedure for the same.

CREATE PROCEDURE get_tree(IN id int)
 BEGIN
 DECLARE child_id int;
 DECLARE prev_id int;
 SET prev_id = id;
 SET child_id=0;
 SELECT col3 into child_id 
 FROM table1 WHERE col1=id ;
 create TEMPORARY  table IF NOT EXISTS temp_table as (select * from table1 where 1=0);
 truncate table temp_table;
 WHILE child_id <> 0 DO
   insert into temp_table select * from table1 WHERE col1=prev_id;
   SET prev_id = child_id;
   SET child_id=0;
   SELECT col3 into child_id
   FROM TABLE1 WHERE col1=prev_id;
 END WHILE;
 select * from temp_table;
 END //

We are using temp table to store results of the output and as the temp tables are session based we wont there will be not be any issue regarding output data being incorrect.

SQL FIDDLE Demo





Try this query:

SELECT 
    col1, col2, @pv := col3 as 'col3' 
FROM 
    table1
JOIN 
    (SELECT @pv := 1) tmp
WHERE 
    col1 = @pv

SQL FIDDLE Demo:

| COL1 | COL2 | COL3 |
+------+------+------+
|    1 |    a |    5 |
|    5 |    d |    3 |
|    3 |    k |    7 |

Note
parent_id value should be less than the child_id for this solution to work.

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