SnowflakeDocumentation Home

HAVING clause

Important Notice: Migration of Documentation Website

Please be advised that our documentation website is currently undergoing a migration to a new platform. To ensure you have access to the most up-to-date information, we kindly request that you visit our new documentation website located at:

Official Snowflake Snowconvert Documentation

For any immediate assistance or if you encounter any issues, please contact our support team at snowconvert-support@snowflake.com.

Thank you for your understanding.

Description

The HAVING clause applies a condition to the intermediate grouped result set that a query returns. (Redshift SQL Language Reference HAVING Clause)

The HAVING clause is fully supported in Snowflake.

Grammar Syntax

[ HAVING condition ]

Sample Source Patterns

Input Code:

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);

SELECT manager_id, COUNT(id) AS total_employees
INTO manager_employees
FROM
employee
GROUP BY manager_id
HAVING COUNT(id) > 2
ORDER BY manager_id;

Output Code:

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": "02/06/2025",  "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);

CREATE TABLE IF NOT EXISTS manager_employees AS
  SELECT manager_id, COUNT(id) AS total_employees
  FROM
    employee
  GROUP BY manager_id
  HAVING COUNT(id) > 2
  ORDER BY manager_id;

Known Issues

There are no known issues.

Related EWIs.

There are no related EWIs.

PreviousGROUP BY clauseNextUNION, INTERSECT, and EXCEPT

Last updated