The FROM clause in a query lists the table references (tables, views, and subqueries) that data is selected from. If multiple table references are listed, the tables must be joined, using appropriate syntax in either the FROM clause or the WHERE clause. If no join criteria are specified, the system processes the query as a cross-join. ()
The is partially supported in Snowflake. is not currently supported.
Grammar Syntax
FROM table_reference [, ...]
<table_reference> ::=
with_subquery_table_name [ table_alias ]
table_name [ * ] [ table_alias ]
( subquery ) [ table_alias ]
table_reference [ NATURAL ] join_type table_reference
[ ON join_condition | USING ( join_column [, ...] ) ]
table_reference PIVOT (
aggregate(expr) [ [ AS ] aggregate_alias ]
FOR column_name IN ( expression [ AS ] in_alias [, ...] )
) [ table_alias ]
table_reference UNPIVOT [ INCLUDE NULLS | EXCLUDE NULLS ] (
value_column_name
FOR name_column_name IN ( column_reference [ [ AS ]
in_alias ] [, ...] )
) [ table_alias ]
UNPIVOT expression AS value_alias [ AT attribute_alias ]
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);
CREATE TABLE department (
id INT,
name VARCHAR(50),
manager_id INT
);
INSERT INTO department(id, name, manager_id) VALUES
(1, 'HR', 100),
(2, 'Sales', 101),
(3, 'Engineering', 102),
(4, 'Marketing', 103);
SELECT e.name AS employee_name, d.name AS department_name
INTO employees_in_department
FROM employee e
INNER JOIN department d ON e.manager_id = d.manager_id;
EMPLOYEE_NAME
DEPARTMENT_NAME
John
HR
Jorge
Sales
Kwaku
Sales
Liu
Sales
Mateo
Engineering
Nikki
Marketing
Paulo
Marketing
Richard
Marketing
Sofía
Engineering
Output Code:
IN -> 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 department (
id INT,
name VARCHAR(50),
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 department (id, name, manager_id) VALUES
(1, 'HR', 100),
(2, 'Sales', 101),
(3, 'Engineering', 102),
(4, 'Marketing', 103);
CREATE TABLE IF NOT EXISTS employees_in_department AS
SELECT e.name AS employee_name, d.name AS department_name
FROM
employee e
INNER JOIN
department d ON e.manager_id = d.manager_id;
