WITH clause

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Description

A WITH clause is an optional clause that precedes the SELECT list in a query. The WITH clause defines one or more common_table_expressions. Each common table expression (CTE) defines a temporary table, which is similar to a view definition. You can reference these temporary tables in the FROM clause. (Redshift SQL Language Reference WITH Clause)

The WITH clause is fully supported in Snowflake.

Grammar Syntax

[ WITH [RECURSIVE] common_table_expression [, common_table_expression , ...] ]

--Where common_table_expression can be either non-recursive or recursive. 
--Following is the non-recursive form:
CTE_table_name [ ( column_name [, ...] ) ] AS ( query )

--Following is the recursive form of common_table_expression:
CTE_table_name (column_name [, ...] ) AS ( recursive_query )

Sample Source Patterns

Recursive form

Input Code:

Redshift

IN -> Redshift_01.sql

CREATE TABLE employee (
  id INT,
  name VARCHAR(20),
  manager_id INT
);
  
INSERT INTO employee(id, name, manager_id) VALUES
(100, 'Carlos', null),
(101, 'John', 100),
(102, 'Jorge', 101),
(103, 'Kwaku', 101),
(110, 'Liu', 101),
(106, 'Mateo', 102),
(110, 'Nikki', 103),
(104, 'Paulo', 103),
(105, 'Richard', 103),
(120, 'Saanvi', 104),
(200, 'Shirley', 104),
(201, 'Sofía', 102),
(205, 'Zhang', 104);
  

WITH RECURSIVE john_org(id, name, manager_id, level) AS
( SELECT id, name, manager_id, 1 AS level
  FROM employee
  WHERE name = 'John'
  UNION ALL
  SELECT e.id, e.name, e.manager_id, level + 1 AS next_level
  FROM employee e, john_org j
  WHERE e.manager_id = j.id and level < 4
)
SELECT DISTINCT id, name, manager_id FROM john_org ORDER BY manager_id;

Results

Output Code:

Snowflake

OUT -> Redshift_01.sql

CREATE TABLE employee (
  id INT,
  name VARCHAR(20),
  manager_id INT
)
COMMENT = '{ "origin": "sf_sc", "name": "snowconvert", "version": {  "major": 0,  "minor": 0,  "patch": "0" }, "attributes": {  "component": "redshift",  "convertedOn": "07/11/2025",  "domain": "no-domain-provided" }}';

INSERT INTO employee (id, name, manager_id) VALUES
(100, 'Carlos', null),
(101, 'John', 100),
(102, 'Jorge', 101),
(103, 'Kwaku', 101),
(110, 'Liu', 101),
(106, 'Mateo', 102),
(110, 'Nikki', 103),
(104, 'Paulo', 103),
(105, 'Richard', 103),
(120, 'Saanvi', 104),
(200, 'Shirley', 104),
(201, 'Sofía', 102),
(205, 'Zhang', 104);


WITH RECURSIVE john_org(id, name, manager_id, level) AS
( SELECT id, name, manager_id, 1 AS level
  FROM
    employee
  WHERE name = 'John'
  UNION ALL
  SELECT e.id, e.name, e.manager_id, level + 1 AS next_level
  FROM
    employee e,
    john_org j
  WHERE e.manager_id = j.id and level < 4
)
SELECT DISTINCT id, name, manager_id FROM
  john_org
ORDER BY manager_id;

Results

Non recursive form

Input Code:

Redshift

IN -> Redshift_02.sql

WITH ManagerHierarchy AS (
    SELECT id AS employee_id, name AS employee_name, manager_id
    FROM employee
)
SELECT e.employee_name AS employee, m.employee_name AS manager
FROM ManagerHierarchy e
LEFT JOIN ManagerHierarchy m ON e.manager_id = m.employee_id;

Results

Output Code:

Snowflake

OUT -> Redshift_02.sql

WITH ManagerHierarchy AS (
    SELECT id AS employee_id, name AS employee_name, manager_id
    FROM
    employee
)
SELECT e.employee_name AS employee, m.employee_name AS manager
FROM
    ManagerHierarchy e
LEFT JOIN
    ManagerHierarchy m ON e.manager_id = m.employee_id;

Results

Related EWIs

There are no known issues.