SnowConvert: Redshift SELECT INTO Statement¶

SELECT INTO¶

Description¶

Returns rows from tables, views, and user-defined functions and inserts them into a new table. (Redshift SQL Language Reference SELECT statement)

Grammar Syntax¶

 [ WITH with_subquery [, ...] ]
SELECT
[ TOP number ] [ ALL | DISTINCT ]
* | expression [ AS output_name ] [, ...]
INTO [ TEMPORARY | TEMP ] [ TABLE ] new_table
[ FROM table_reference [, ...] ]
[ WHERE condition ]
[ GROUP BY expression [, ...] ]
[ HAVING condition [, ...] ]
[ { UNION | INTERSECT | { EXCEPT | MINUS } } [ ALL ] query ]
[ ORDER BY expression
[ ASC | DESC ]
[ LIMIT { number | ALL } ]
[ OFFSET start ]

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For more information please refer to each of the following links:

WITH clause
SELECT list
FROM clause
WHERE clause
CONNECT BY clause
GROUP BY clause
HAVING clause
QUALIFY clause
UNION, INTERSECT, and EXCEPT
ORDER BY clause
LIMIT and OFFSET clauses
Local Variables and Parameters

FROM clause¶

Description¶

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. (Redshift SQL Language Reference FROM Clause)

Warning

The FROM clause is partially supported in Snowflake. Object unpivoting 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 ]

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Sample Source Patterns¶

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);

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;

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EMPLOYEE_NAME	DEPARTMENT_NAME
John	HR
Jorge	Sales
Kwaku	Sales
Liu	Sales
Mateo	Engineering
Nikki	Marketing
Paulo	Marketing
Richard	Marketing
Sofía	Engineering

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": "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;

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EMPLOYEE_NAME	DEPARTMENT_NAME
John	HR
Jorge	Sales
Kwaku	Sales
Liu	Sales
Mateo	Engineering
Nikki	Marketing
Paulo	Marketing
Richard	Marketing
Sofía	Engineering

Known Issues¶

There are no known issues.

GROUP BY clause¶

Description¶

The GROUP BY clause identifies the grouping columns for the query. Grouping columns must be declared when the query computes aggregates with standard functions such as SUM, AVG, and COUNT. (Redshift SQL Language Reference GROUP BY Clause)

The GROUP BY clause is fully supported in Snowflake.

Grammar Syntax

 GROUP BY expression [, ...]

Copy

Sample Source Patterns

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);

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

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MANAGER_ID	TOTAL_EMPLOYEES
100	1
101	3
102	2
103	3
104	3
	1

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": "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
  ORDER BY manager_id;

Copy

MANAGER_ID	TOTAL_EMPLOYEES
100	1
101	3
102	2
103	3
104	3
	1

Known Issues

There are no known issues.

Related EWIs.

There are no related EWIs.

HAVING clause

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 ]

Copy

Sample Source Patterns¶

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);

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;

Copy

MANAGER_ID	TOTAL_EMPLOYEES
101	3
103	3
104	3

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": "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;

Copy

MANAGER_ID	TOTAL_EMPLOYEES
101	3
103	3
104	3

Known Issues¶

There are no known issues.

Related EWIs.¶

There are no related EWIs.

LIMIT and OFFSET clauses¶

Description¶

The LIMIT and OFFSET clauses retrieves and skips the number of rows specified in the number.

The LIMIT and OFFSET clauses are fully supported in Snowflake.

Grammar Syntax

 [ LIMIT { number | ALL } ]
[ OFFSET start ]

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Sample Source Patterns

LIMIT number

Input Code:

 SELECT id, name, manager_id, salary
INTO limited_employees
FROM employee
LIMIT 5;

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ID	NAME	MANAGER_ID	SALARY
100	Carlos		120000.00
101	John	100	90000.00
102	Jorge	101	95000.00
103	Kwaku	101	105000.00
104	Paulo	102	110000.00

Output Code:

 CREATE TABLE IF NOT EXISTS limited_employees AS
SELECT id, name, manager_id, salary
FROM
employee
LIMIT 5;

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ID	NAME	MANAGER_ID	SALARY
100	Carlos		120000.00
101	John	100	90000.00
102	Jorge	101	95000.00
103	Kwaku	101	105000.00
104	Paulo	102	110000.00

LIMIT ALL

Input Code:

 SELECT id, name, manager_id, salary
INTO limited_employees
FROM employee
LIMIT ALL;

Copy

ID	NAME	MANAGER_ID	SALARY
100	Carlos		120000.00
101	John	100	90000.00
102	Jorge	101	95000.00
103	Kwaku	101	105000.00
104	Paulo	102	110000.00
105	Richard	102	85000.00
106	Mateo	103	95000.00
107	Liu	103	108000.00
108	Zhang	104	95000.00

Output Code:

 CREATE TABLE IF NOT EXISTS limited_employees AS
SELECT id, name, manager_id, salary
FROM
employee
LIMIT NULL;

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ID	NAME	MANAGER_ID	SALARY
100	Carlos		120000.00
101	John	100	90000.00
102	Jorge	101	95000.00
103	Kwaku	101	105000.00
104	Paulo	102	110000.00
105	Richard	102	85000.00
106	Mateo	103	95000.00
107	Liu	103	108000.00
108	Zhang	104	95000.00

OFFSET without LIMIT

Snowflake doesn’t support OFFSET without LIMIT. The LIMIT is added after transformation with NULL, which is the default LIMIT.

Input Code:

 SELECT id, name, manager_id, salary
INTO limited_employees
FROM employee
OFFSET 5;

Copy

ID	NAME	MANAGER_ID	SALARY
105	Richard	102	85000.00
106	Mateo	103	95000.00
107	Liu	103	108000.00
108	Zhang	104	95000.00

Output Code:

 CREATE TABLE IF NOT EXISTS limited_employees AS
SELECT id, name, manager_id, salary
FROM
employee
LIMIT NULL
OFFSET 5;

Copy

ID	NAME	MANAGER_ID	SALARY
105	Richard	102	85000.00
106	Mateo	103	95000.00
107	Liu	103	108000.00
108	Zhang	104	95000.00

Known Issues

There are no known issues.

Related EWIs.

There are no related EWIs.

Local Variables and Parameters

Description

Redshift also allow to SELECT INTO variables when the statement is executed inside stored procedures.

This pattern is fully supported in Snowflake.

Grammar Syntax¶

 SELECT [ select_expressions ] INTO target [ select_expressions ] FROM ...;

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Sample Source Patterns¶

SELECT INTO with expressions at the left¶

Input Code:¶

 CREATE OR REPLACE PROCEDURE test_sp1(out param1 int)
AS $$
DECLARE
    var1 int;
BEGIN
     select 10, 100 into param1, var1;
END;
$$ LANGUAGE plpgsql;

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param1
10

Output Code:¶

 CREATE OR REPLACE PROCEDURE test_sp1 (param1 int)
RETURNS VARIANT
LANGUAGE SQL
COMMENT = '{ "origin": "sf_sc", "name": "snowconvert", "version": {  "major": 0,  "minor": 0,  "patch": "0" }, "attributes": {  "component": "redshift",  "convertedOn": "03/03/2025",  "domain": "test" }}'
AS $$
        DECLARE
            var1 int;
BEGIN
     select 10, 100 into
                : param1,
                : var1;
            RETURN OBJECT_CONSTRUCT('param1', :param1);
END;
$$;

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TEST_SP1
{ “param1”: 10 }

SELECT INTO with expressions at the right¶

Input Code:¶

 CREATE OR REPLACE PROCEDURE test_sp1(out param1 int)
AS $$
DECLARE
    var1 int;
BEGIN
     select into param1, var1 10, 100;
END;
$$ LANGUAGE plpgsql;

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param1
10

Output Code:¶

Since Snowflake doesn’t support this grammar for SELECT INTO, the expressions are moved to the left of the INTO.

 CREATE OR REPLACE PROCEDURE test_sp1 (param1 int)
RETURNS VARIANT
LANGUAGE SQL
COMMENT = '{ "origin": "sf_sc", "name": "snowconvert", "version": {  "major": 0,  "minor": 0,  "patch": "0" }, "attributes": {  "component": "redshift",  "convertedOn": "03/03/2025",  "domain": "test" }}'
AS $$
        DECLARE
            var1 int;
BEGIN
     select
                10, 100
            into
                : param1,
                : var1;
            RETURN OBJECT_CONSTRUCT('param1', :param1);
END;
$$;

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TEST_SP1
{ “param1”: 10 }

Known Issues¶

There are no known issues.

Related EWIs.¶

There are no related EWIs.

ORDER BY clause¶

Description¶

The ORDER BY clause sorts the result set of a query. (Redshift SQL Language Reference Order By Clause)

The ORDER BY clause is fully supported in Snowflake.

Grammar Syntax

 [ ORDER BY expression [ ASC | DESC ] ]
[ NULLS FIRST | NULLS LAST ]
[ LIMIT { count | ALL } ]
[ OFFSET start ]

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Sample Source Patterns

Input Code:

 CREATE TABLE employee (
    id INT,
    name VARCHAR(20),
    manager_id INT,
    salary DECIMAL(10, 2)
);

INSERT INTO employee (id, name, manager_id, salary) VALUES
(100, 'Carlos', NULL, 120000.00),
(101, 'John', 100, 90000.00),
(102, 'Jorge', 101, 95000.00),
(103, 'Kwaku', 101, 105000.00),
(104, 'Paulo', 102, 110000.00),
(105, 'Richard', 102, 85000.00),
(106, 'Mateo', 103, 95000.00),
(107, 'Liu', 103, 108000.00),
(108, 'Zhang', 104, 95000.00);

SELECT id, name, manager_id, salary
INTO salaries
FROM employee
ORDER BY salary DESC NULLS LAST, name ASC NULLS FIRST
LIMIT 5                                        
OFFSET 2;

Copy

ID	NAME	MANAGER_ID	SALARY
107	Liu	103	108000.00
103	Kwaku	101	105000.00
102	Jorge	101	95000.00
106	Mateo	103	95000.00
108	Zhang	104	95000.00

Output Code:

 CREATE TABLE employee (
    id INT,
    name VARCHAR(20),
    manager_id INT,
    salary DECIMAL(10, 2)
)
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, salary) VALUES
(100, 'Carlos', NULL, 120000.00),
(101, 'John', 100, 90000.00),
(102, 'Jorge', 101, 95000.00),
(103, 'Kwaku', 101, 105000.00),
(104, 'Paulo', 102, 110000.00),
(105, 'Richard', 102, 85000.00),
(106, 'Mateo', 103, 95000.00),
(107, 'Liu', 103, 108000.00),
(108, 'Zhang', 104, 95000.00);

CREATE TABLE IF NOT EXISTS salaries AS
    SELECT id, name, manager_id, salary
    FROM
        employee
    ORDER BY salary DESC NULLS LAST, name ASC NULLS FIRST
    LIMIT 5
    OFFSET 2;

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ID	NAME	MANAGER_ID	SALARY
107	Liu	103	108000.00
103	Kwaku	101	105000.00
102	Jorge	101	95000.00
106	Mateo	103	95000.00
108	Zhang	104	95000.00

Known Issues

There are no known issues.

Related EWIs.

There are no related EWIs.

SELECT list

Description

The SELECT list names the columns, functions, and expressions that you want the query to return. The list represents the output of the query. (Redshift SQL Language Reference SELECT list)

The query start options are fully supported in Snowflake. Just keep in mind that in Snowflake the DISTINCT and ALL options must go at the beginning of the query.

Note

In Redshift, if your application allows foreign keys or invalid primary keys, it can cause queries to return incorrect results. For example, a SELECT DISTINCT query could return duplicate rows if the primary key column does not contain all unique values. (Redshift SQL Language Reference SELECT list)

Grammar Syntax¶

 SELECT
[ TOP number ]
[ ALL | DISTINCT ] * | expression [ AS column_alias ] [, ...]

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Sample Source Patterns¶

Top clause¶

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);
  
SELECT TOP 5 id, name, manager_id 
INTO top_employees
FROM employee;

SELECT * FROM top_employees;

Copy

ID	NAME	MANAGER_ID
100	Carlos	null
101	John	100
102	Jorge	101
103	Kwaku	101
110	Liu	101

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": "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 top_employees AS
SELECT TOP 5 id, name, manager_id
  FROM
    employee;

SELECT * FROM
  top_employees;

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ID	NAME	MANAGER_ID
100	Carlos	null
101	John	100
102	Jorge	101
103	Kwaku	101
110	Liu	101

ALL¶

Input Code:¶

SELECT ALL manager_id
INTO manager
FROM employee;

Copy

MANAGER_ID
null
100
101
101
101
102
103
103
103
104
104
102
104

Output Code:¶

 CREATE TABLE IF NOT EXISTS manager AS
SELECT ALL manager_id
FROM
employee;

Copy

MANAGER_ID
null
100
101
101
101
102
103
103
103
104
104
102
104

DISTINCT¶

Input Code:¶

SELECT DISTINCT manager_id
INTO manager
FROM employee;

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MANAGER_ID
null
100
101
102
103
104

Output Code:¶

 CREATE TABLE IF NOT EXISTS manager AS
SELECT DISTINCT manager_id
FROM
employee;

Copy

MANAGER_ID
null
100
101
102
103
104

Known Issues¶

There are no known issues.

Related EWIs.¶

There are no related EWIs.

UNION, INTERSECT, and EXCEPT¶

Description¶

The UNION, INTERSECT, and EXCEPT set operators are used to compare and merge the results of two separate query expressions. (Redshift SQL Language Reference Set Operators)

Set operators are fully supported in Snowflake.

Grammar Syntax

 query
{ UNION [ ALL ] | INTERSECT | EXCEPT | MINUS }
query

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Sample Source Patterns

Input Code:

 SELECT id, name, manager_id
INTO some_employees
FROM
employee
WHERE manager_id = 101

UNION

SELECT id, name, manager_id
FROM
employee
WHERE manager_id = 102

UNION ALL

SELECT id, name, manager_id
FROM
employee
WHERE manager_id = 101

INTERSECT

SELECT id, name, manager_id
FROM
employee
WHERE manager_id = 103

EXCEPT

SELECT id, name, manager_id
FROM
employee
WHERE manager_id = 104;

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ID	NAME	MANAGER_ID
103	Kwaku	101
110	Liu	101
102	Jorge	101
106	Mateo	102
201	Sofía	102

Output Code:

 CREATE TABLE IF NOT EXISTS some_employees AS
SELECT id, name, manager_id
FROM
employee
WHERE manager_id = 101

UNION

SELECT id, name, manager_id
FROM
employee
WHERE manager_id = 102

UNION ALL

SELECT id, name, manager_id
FROM
employee
WHERE manager_id = 101

INTERSECT

SELECT id, name, manager_id
FROM
employee
WHERE manager_id = 103

EXCEPT

SELECT id, name, manager_id
FROM
employee
WHERE manager_id = 104;

Copy

ID	NAME	MANAGER_ID
102	Jorge	101
103	Kwaku	101
110	Liu	101
106	Mateo	102
201	Sofía	102

Known Issues

There are no known issues.

Related EWIs.

There are no related EWIs.

WHERE clause

Description

The WHERE clause contains conditions that either join tables or apply predicates to columns in tables. (Redshift SQL Language Reference WHERE Clause)

The WHERE clause is fully supported in Snowflake.

Grammar Syntax¶

 [ WHERE condition ]

Copy

Sample Source Patterns¶

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);

SELECT id, name, manager_id
INTO employee_names
FROM employee
WHERE name LIKE 'J%';

Copy

ID	NAME	MANAGER_ID
101	John	100
102	Jorge	101

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": "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 employee_names AS
  SELECT id, name, manager_id
  FROM
    employee
  WHERE name LIKE 'J%' ESCAPE '\\';

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ID	NAME	MANAGER_ID
101	John	100
102	Jorge	101

Known Issues¶

There are no known issues.

Related EWIs.¶

There are no related EWIs.

WITH clause¶

Description¶

A WITH clause is an optional clause that precedes the SELECT INTO 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 )

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Sample Source Patterns

Non-Recursive form

Input Code:

 CREATE TABLE orders (
    order_id INT,
    customer_id INT,
    order_date DATE,
    total_amount DECIMAL(10,2)
);


INSERT INTO orders (order_id, customer_id, order_date, total_amount)
VALUES
(1, 101, '2024-02-01', 250.00),
(2, 102, '2024-02-02', 600.00),
(3, 103, '2024-02-03', 150.00),
(4, 104, '2024-02-04', 750.00),
(5, 105, '2024-02-05', 900.00);


WITH HighValueOrders AS (
    SELECT
        order_id,
        customer_id,
        order_date,
        total_amount
    FROM orders
    WHERE total_amount > 500
)
SELECT * INTO high_value_orders FROM HighValueOrders;

SELECT * FROM high_value_orders;

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ORDER_ID	CUSTOMER_ID	ORDER_DATE	TOTAL_AMOUNT
2	102	2024-02-02	600.00
4	104	2024-02-04	750.00
5	105	2024-02-05	900.00

Output Code:

 CREATE TABLE orders (
    order_id INT,
    customer_id INT,
    order_date DATE,
    total_amount DECIMAL(10,2)
)
COMMENT = '{ "origin": "sf_sc", "name": "snowconvert", "version": {  "major": 0,  "minor": 0,  "patch": "0" }, "attributes": {  "component": "redshift",  "convertedOn": "03/03/2025",  "domain": "test" }}';


INSERT INTO orders (order_id, customer_id, order_date, total_amount)
VALUES
(1, 101, '2024-02-01', 250.00),
(2, 102, '2024-02-02', 600.00),
(3, 103, '2024-02-03', 150.00),
(4, 104, '2024-02-04', 750.00),
(5, 105, '2024-02-05', 900.00);

CREATE TABLE IF NOT EXISTS high_value_orders AS
WITH HighValueOrders AS (
    SELECT
        order_id,
        customer_id,
        order_date,
        total_amount
    FROM
        orders
    WHERE total_amount > 500
    )
    SELECT *
    FROM
    HighValueOrders;
    
SELECT * FROM
    high_value_orders;

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ORDER_ID	CUSTOMER_ID	ORDER_DATE	TOTAL_AMOUNT
2	102	2024-02-02	600.00
4	104	2024-02-04	750.00
5	105	2024-02-05	900.00

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 into new_org FROM john_org ORDER BY manager_id;

SELECT * FROM new_org;

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ID	NAME	MANAGER_ID
101	John	100
103	Kwaku	101
102	Jorge	101
110	Liu	101
106	Mateo	102
201	Sofía	102
105	Richard	103
110	Nikki	103
104	Paulo	103
120	Saanvi	104
200	Shirley	104
205	Zhang	104

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": "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 new_org AS
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;

   
SELECT * FROM
   new_org;

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ID	NAME	MANAGER_ID
101	John	100
103	Kwaku	101
102	Jorge	101
110	Liu	101
106	Mateo	102
201	Sofía	102
105	Richard	103
110	Nikki	103
104	Paulo	103
120	Saanvi	104
200	Shirley	104
205	Zhang	104

Known Issues

There are no known issues.

Related EWIs.

There are no related EWIs.