WITH clause
Description
A
WITH
clause is an optional clause that precedes the SELECT list in a query. TheWITH
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 theFROM
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:
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;
Output Code:
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": "11/05/2024", "domain": "test" }}';
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
RTRIM( name) = RTRIM( '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;
Non recursive form
Input Code:
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;
Output Code:
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;
Related EWIs
There are no known issues.
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