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Franck Pachot for YugabyteDB

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How to write SQL recursive CTE in 5 steps

You don't need a Graph database to follow a hierarchy of nodes. SQL can do joins, self-join, and even joins to its previous result, iteratively. This WITH RECURSIVE clause is often called "Recursive CTE". This might require a bit of abstraction for the developers used to procedural languages. Here is how I explain it by joining "manually" the two first levels, without recursion, and then switching to "recursive" to go further. Starting with the first two levels helps to validate the logic easily.

(0) Example: create the EMP table

I'll create the oldest sample table: EMP for employees with a self-reference to the manager:

CREATE TABLE emp (empno, ename,    job,        mgr,   hiredate,     sal, comm, deptno, email, other_info) as values
(7369, 'SMITH',  'CLERK',     7902, '1980-12-17',  800, NULL,   20,'SMITH@acme.com', '{"skills":["accounting"]}'),
(7499, 'ALLEN',  'SALESMAN',  7698, '1981-02-20', 1600,  300,   30,'ALLEN@acme.com', null),
(7521, 'WARD',   'SALESMAN',  7698, '1981-02-22', 1250,  500,   30,'WARD@compuserve.com', null),
(7566, 'JONES',  'MANAGER',   7839, '1981-04-02', 2975, NULL,   20,'JONES@gmail.com', null),
(7654, 'MARTIN', 'SALESMAN',  7698, '1981-09-28', 1250, 1400,   30,'MARTIN@acme.com', null),
(7698, 'BLAKE',  'MANAGER',   7839, '1981-05-01', 2850, NULL,   30,'BLAKE@hotmail.com', null),
(7782, 'CLARK',  'MANAGER',   7839, '1981-06-09', 2450, NULL,   10,'CLARK@acme.com', '{"skills":["C","C++","SQL"]}'),
(7788, 'SCOTT',  'ANALYST',   7566, '1982-12-09', 3000, NULL,   20,'SCOTT@acme.com', '{"cat":"tiger"}'),
(7839, 'KING',   'PRESIDENT', NULL, '1981-11-17', 5000, NULL,   10,'KING@aol.com', null),
(7844, 'TURNER', 'SALESMAN',  7698, '1981-09-08', 1500,    0,   30,'TURNER@acme.com', null),
(7876, 'ADAMS',  'CLERK',     7788, '1983-01-12', 1100, NULL,   20,'ADAMS@acme.org', null),
(7900, 'JAMES',  'CLERK',     7698, '1981-12-03',  950, NULL,   30,'JAMES@acme.org', null),
(7902, 'FORD',   'ANALYST',   7566, '1981-12-03', 3000, NULL,   20,'FORD@acme.com', '{"skills":["SQL","CQL"]}'),
(7934, 'MILLER', 'CLERK',     7782, '1982-01-23', 1300, NULL,   10,'MILLER@acme.com', null)
;

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The self reference from mgr to empno can be declared as:

ALTER TABLE emp ADD PRIMARY KEY (empno)
;
ALTER TABLE emp ADD FOREIGN KEY (mgr) REFERENCES emp (empno)
;
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This is not needed but it ensures the correctness of that that you query so that you can focus on your business logic rather than consistency test.

(1) find the root (level 0)

The first step is to identify the root of the graph. Do you want to start by the employees and show their manager, or start by the mangager and show its employees? I'll do the latter. The root of the hierarchy is the one with no manager (where mgr is null)

-- (1) query to find the root
select emp.empno,emp.ename,emp.mgr
from emp where mgr is null           -- (1) this is the level 0
;
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The result is simple:

yugabyte-> ;

 empno | ename | mgr
-------+-------+-----
  7839 | KING  |
(1 row)

yugabyte=>
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(2) define this level 0 in the WITH clause

I'll put it in a CTE (Common Table Expression), labelled as levelN, but until we add recursion it is level 0, and I select from it:

with
-- (2) define the root as as levelN (which is level 0 until we add recursion)
levelN as (
-- (1) query to find the root
select emp.empno,emp.ename,emp.mgr
from emp where mgr is null           -- (1) this is the level 0
) 
-- (2) query this level
select * from levelN
;
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The result is the same. The top manager is KING. We just named it and put it in a WITH clause to prepare for future recursion.

yugabyte-> ;

 empno | ename | mgr
-------+-------+-----
  7839 | KING  |
(1 row)

yugabyte=>
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I'm just building the blocks one by one. And I've left the comments to make it clear what is taken from the previous step and what is added.

(3) define the join to the next level

Now that my level 0 is defined, I query the emp table again, to get the next level, with a join to level 0 to get the manager name:

with 
-- (2) define the root as as levelN which is level 0 until we add recursion
levelN as (
-- (1) query to find the root
select emp.empno,emp.ename,emp.mgr
from emp where mgr is null           -- (1) this is the level 0
) 
-- (3) join employees with levelN to get level N+1 (which is level 1 here)
select emp.empno,emp.ename,emp.mgr,mgr.ename mgr_name
from emp
join levelN mgr on mgr.empno=emp.mgr
;
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I have aliased the level 0 as mgr because it will be the manager of this new level of employee, for which I keep the table name emp. Then the join on mgr.empno=emp.mgr links the two levels.

In the join result, I have added mgr.ename as mgr_name in order to see the name of the manager in the same line.

So here here are the employees reporting to KING:

yugabyte-> ;

 empno | ename | mgr  | mgr_name
-------+-------+------+----------
  7698 | BLAKE | 7839 | KING
  7566 | JONES | 7839 | KING
  7782 | CLARK | 7839 | KING
(3 rows)

yugabyte=>
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(4) concatenate the two levels with UNION

Now that I have two levels, I can start to build the hierarchy by concatenating them with a UNION ALL. You see levelN, wich is the level 0, in the first branch of the UNION, to show the level 0 and in the second branch to join the two levels.

Because I have added the manager name to the level 1, I add it also in the level zero, as an empty string here, in order to have the same structure in the UNION.

with 
-- (2) define the root as as levelN which is level0 until we add recursion
levelN as (
-- (1) query to find the root      --(4) add all columns
select emp.empno,emp.ename,emp.mgr,'' as mgr
from emp where mgr is null           -- (1) this is the level 0
) 
-- (4) concatenate level 0 on top of level l
select * from levelN  
union all
-- (3) join employees with level n to get level n+1 (which is level 1 here)
select emp.empno,emp.ename,emp.mgr,mgr.ename mgr_name
from emp
join levelN mgr on mgr.empno=emp.mgr
;
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This starts to look good, here are the two levels together:

yugabyte-> ;

 empno | ename | mgr  | mgr_name
-------+-------+------+----------
  7839 | KING  |      |
  7698 | BLAKE | 7839 | KING
  7566 | JONES | 7839 | KING
  7782 | CLARK | 7839 | KING
(4 rows)

yugabyte=>
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This is where you can validate that your join is correct.

The next step will just add recursion so that you don't need to do the same to add the level 3. The same definition (UNION ALL and JOIN from the previous level) can get to the next levels.

(5) get it in one recursive WITH clause

Now that I have validated the result for the two first levels, it is time to get it recursive by adding the RECURSIVE keyword and moving the UNION ALL into the Common Table Expression. The first part of the UNION ALL is executed only once in the first iteration, but the second part will concatenate the next level.

-- (5) change to recursive 
with recursive
-- (2) define the root as as levelN which is level0 until we add recursion
levelN as (
-- (1) query to find the root      --(4) add all columns
select emp.empno,emp.ename,emp.mgr,'' as mgr
from emp where mgr is null           -- (1) this is the level 0
-- (5) move second branch into the CTE which is the one that will be iterated. It references levelN  from the previous iteration.
union all
-- (3) join employees with level n to get level n+1 (which is level 1 here)
select emp.empno,emp.ename,emp.mgr,mgr.ename mgr_name
from emp
join levelN mgr on mgr.empno=emp.mgr
) 
-- (4) concatenate level 0 on top of level l
select * from levelN
-- (5) we query only from levelN here, which includes the previous levels thanks to the recursion  
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If you have validated the join with the first two level, this last transformation is automatic. At execution time, the first iteration will get the first level and the second part of the UNION ALL will have no rows because of the join with the previous result, which is empty (there is no previous level). The next iteration will add the result of the join. And again until the join returns no rows because we are at a leaf, with the previous result, taken as a list of managers, has no employees.

yugabyte-> ;

 empno | ename  | mgr  | mgr_name
-------+--------+------+----------
  7839 | KING   |      |
  7698 | BLAKE  | 7839 | KING
  7566 | JONES  | 7839 | KING
  7782 | CLARK  | 7839 | KING
  7788 | SCOTT  | 7566 | JONES
  7902 | FORD   | 7566 | JONES
  7844 | TURNER | 7698 | BLAKE
  7499 | ALLEN  | 7698 | BLAKE
  7521 | WARD   | 7698 | BLAKE
  7654 | MARTIN | 7698 | BLAKE
  7900 | JAMES  | 7698 | BLAKE
  7934 | MILLER | 7782 | CLARK
  7876 | ADAMS  | 7788 | SCOTT
  7369 | SMITH  | 7902 | FORD
(14 rows)

yugabyte=>
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This is where the magic is. If you carefully test your first two levels, all others will come by transforming the query to be recursive, iterating by joining on its previous result.
In the execution plan, you see it as Recursive union

                                                              QUERY PLAN
--------------------------------------------------------------------------------------------------------------------------------------
 CTE Scan on leveln  (cost=20.23..22.86 rows=131 width=72) (actual time=0.023..0.281 rows=14 loops=1)
   CTE leveln
     ->  Recursive Union  (cost=0.00..20.23 rows=131 width=46) (actual time=0.021..0.269 rows=14 loops=1)
           ->  Seq Scan on emp  (cost=0.00..1.14 rows=1 width=46) (actual time=0.019..0.021 rows=1 loops=1)
                 Filter: (mgr IS NULL)
           ->  Hash Join  (cost=0.33..1.65 rows=13 width=46) (actual time=0.054..0.058 rows=3 loops=4)
                 Hash Cond: (emp_1.mgr = mgr.empno)
                 ->  Seq Scan on emp emp_1  (cost=0.00..1.14 rows=14 width=14) (actual time=0.002..0.003 rows=14 loops=4)
                 ->  Hash  (cost=0.20..0.20 rows=10 width=36) (actual time=0.044..0.044 rows=4 loops=4)
                       ->  WorkTable Scan on leveln mgr  (cost=0.00..0.20 rows=10 width=36) (actual time=0.001..0.001 rows=4 loops=4)
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The number of loops tells you how many time the second part was executed. All this runs the same in PostgreSQL and YugabyteDB. Here is the example on db<>fiddle: https://dbfiddle.uk/sChh3Y09

A final remark: the declaration is recursive but the execution is iterative, with a loop until no rows are returned.

Top comments (8)

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geraldew profile image
geraldew

Always good to see articles describing how recursive queries work - disclaimer, I've also written one here on Dev (but it wasn't intended as an introduction to the concept).

With recursion, there's always a need to be sure that the looping will end. I'm not seeing anything here to be certain of that - other than it being known to be true for the example data. There are various possible solutions (I have a my own preferred method) so maybe that would make a good follow-up piece.

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alfiqmiq profile image
Marek • Edited

Always good to include hints in query (example for MySQL):
WITH
...
SELECT /*+MAX_EXECUTION_TIME(1000) SET_VAR(CTE_MAX_RECURSION_DEPTH=100)*/
...

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detzam profile image
webstuff

this one is quite interesting. But its not PL SQL, is it?

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markusnetz89 profile image
Markus Netz

No, it is standard sql

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michaeldoshea profile image
/* Michael D O'Shea */

It is Postgres.

In the Oracle flavour of SQL (not PL/SQL as you wrote, that is something very different) I wrote this up a few years ago strychnine.co.uk/2017/08/postgresq... I rabbit on a bit in there but the difference between Oracle and Postgres at the time was far from dramatic, at least for this functionality.

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franckpachot profile image
Franck Pachot • Edited

Michael, to be fair, Oracle had CONNECT BY from the initial version (seen in 2.3 User Guide from 1981 ๐Ÿค“), many years before WITH RECURSIVE came into the SQL standard. They were late to the ANSI standard but early on the feature

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michaeldoshea profile image
/* Michael D O'Shea */

Yeah I get it Frank, it's like DECODE (now implemented CASE WHEN ... END), NVL and its friends, the (+) for outer joins and so on, and all remain in the implementation as Oracle incorporated some features/functions in-part early on and the ANSI/ISO standards have since "caught up". Oracle hasn't for the most part had language feature breaking changes either, so all this junk DNA remains in their flagship product (well the removal of WM_CONCAT was a breaking change, but it was an undocumented feature as an interim approach to LISTAGG so doesn't really count).

Perhaps another way of looking at all of this is that Oracle were leading the way winning the awards at the beginning, but now a bit of a dinosaur and not winning the awards now. Younger spunky database implementations don't carry the baggage of these early RDBMS implementations. I include Postgres in this category, and its forks (eg. Netezza), that implement CTE's in a more standardised way.

Even under the bonnet Oracle can be a little sad too, as you know. I think it was Oracle 9 (maybe 9.2 before it worked) that Oracle implemented ANSI joins. When you look at a 10046 trace file however, you can see the ANSI SQL is rewritten back to the proprietary format, some sort of internal transpiler shoehorned in Oracle 9, and two semantically equivalent queries (one using ANSI joins and one using (+)), both hinted ..... well the ANSI code often loses the hints in the rewrite and the execution plan doesn't adhere to the "rule" (yeah yeah, hints are rules, we both understand this too).

On the subject of execution hints, and Postgres, well it's just not there and the (non) implementation defended wiki.postgresql.org/wiki/Optimizer... It will be interesting to see how this changes in the future, if it changes. If it changes, someone will be writing a nice blog article on it too. I will keep my eyes peeled :-)

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franckpachot profile image
Franck Pachot

Yes, ANSI joins introduced 20 years ago and still many bugs ๐Ÿคท๐Ÿผโ€โ™‚๏ธ
About pg_hint_plan, at in YugabyteDB we have it by default. I see no reason not to allow easy troubleshooting / workaround.