SnowConvert: Redshift SELECT Statement¶

SELECT¶

Description¶

Returns rows from tables, views, and user-defined functions. (Redshift SQL Language Reference SELECT statement)

Grammar Syntax¶

 [ WITH with_subquery [, ...] ]
SELECT
[ TOP number | [ ALL | DISTINCT ]
* | expression [ AS output_name ] [, ...] ]
[ FROM table_reference [, ...] ]
[ WHERE condition ]
[ [ START WITH expression ] CONNECT BY expression ]
[ GROUP BY expression [, ...] ]
[ HAVING condition ]
[ QUALIFY condition ]
[ { UNION | ALL | INTERSECT | EXCEPT | MINUS } 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

CONNECT BY clause¶

Description¶

The CONNECT BY clause specifies the relationship between rows in a hierarchy. You can use CONNECT BY to select rows in a hierarchical order by joining the table to itself and processing the hierarchical data. (Redshift SQL Language Reference CONNECT BY Clause)

The CONNECT BY clause is supported in Snowflake.

Grammar Syntax

 [START WITH start_with_conditions]
CONNECT BY connect_by_conditions

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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);

SELECT COUNT(*)
FROM
Employee "start"
CONNECT BY PRIOR id = manager_id
START WITH name = 'John';

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COUNT(*)
12

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);

SELECT COUNT(*)
FROM
  Employee "start"
CONNECT BY PRIOR id = manager_id
START WITH RTRIM( name) = RTRIM( 'John');

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COUNT(*)
12

There are no known issues.

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¶

Join types¶

Snowflake supports all types of joins. For more information, see the JOIN documentation.

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
FROM employee e
INNER JOIN department d ON e.manager_id = d.manager_id;

SELECT e.name AS employee_name, d.name AS department_name
FROM employee e
LEFT JOIN department d ON e.manager_id = d.manager_id;

SELECT d.name AS department_name, e.name AS manager_name
FROM department d
RIGHT JOIN employee e ON d.manager_id = e.id;

SELECT e.name AS employee_name, d.name AS department_name
FROM employee e
FULL JOIN department d ON e.manager_id = d.manager_id;

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Inner Join¶

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

Left Join¶

EMPLOYEE_NAME	DEPARTMENT_NAME
Carlos	null
John	HR
Jorge	Sales
Kwaku	Sales
Liu	Sales
Mateo	Engineering
Nikki	Marketing
Paulo	Marketing
Richard	Marketing
Saanvi	null
Shirley	null
Sofía	Engineering
Zhang	null

Right Join¶

DEPARTMENT_NAME	MANAGER_NAME
HR	Carlos
Sales	John
Engineering	Jorge
Marketing	Kwaku
null	Liu
null	Mateo
null	Nikki
null	Paulo
null	Richard
null	Saanvi
null	Shirley
null	Sofía
null	Zhang

Full Join¶

EMPLOYEE_NAME	DEPARTMENT_NAME
Carlos	null
John	HR
Jorge	Sales
Kwaku	Sales
Liu	Sales
Mateo	Engineering
Nikki	Marketing
Paulo	Marketing
Richard	Marketing
Saanvi	null
Shirley	null
Sofía	Engineering
Zhang	null

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);

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": "11/05/2024",  "domain": "test" }}';

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
FROM
employee e
INNER JOIN
  department d ON e.manager_id = d.manager_id;

SELECT e.name AS employee_name, d.name AS department_name
FROM
employee e
LEFT JOIN
  department d ON e.manager_id = d.manager_id;

SELECT d.name AS department_name, e.name AS manager_name
FROM
department d
RIGHT JOIN
  employee e ON d.manager_id = e.id;

SELECT e.name AS employee_name, d.name AS department_name
FROM
employee e
FULL JOIN
  department d ON e.manager_id = d.manager_id;

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Inner Join¶

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

Left Join¶

EMPLOYEE_NAME	DEPARTMENT_NAME
Carlos	null
John	HR
Jorge	Sales
Kwaku	Sales
Liu	Sales
Mateo	Engineering
Nikki	Marketing
Paulo	Marketing
Richard	Marketing
Saanvi	null
Shirley	null
Sofía	Engineering
Zhang	null

Right Join¶

DEPARTMENT_NAME	MANAGER_NAME
HR	Carlos
Sales	John
Engineering	Jorge
Marketing	Kwaku
null	Liu
null	Mateo
null	Nikki
null	Paulo
null	Richard
null	Saanvi
null	Shirley
null	Sofía
null	Zhang

Full Join¶

EMPLOYEE_NAME	DEPARTMENT_NAME
Carlos	null
John	HR
Jorge	Sales
Kwaku	Sales
Liu	Sales
Mateo	Engineering
Nikki	Marketing
Paulo	Marketing
Richard	Marketing
Saanvi	null
Shirley	null
Sofía	Engineering
Zhang	null

Pivot Clause¶

Note

Column aliases cannot be used in the IN clause of the PIVOT query in Snowflake.

Input Code:¶

 SELECT *
FROM
    (SELECT e.manager_id, d.name AS department, e.id AS employee_id
     FROM employee e
     JOIN department d ON e.manager_id = d.manager_id) AS SourceTable
PIVOT
    (
     COUNT(employee_id)
     FOR department IN ('HR', 'Sales', 'Engineering', 'Marketing')
    ) AS PivotTable;

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MANAGER_ID	‘HR’	‘Sales’	‘Engineering’	‘Marketing’
100	1	0	0	0
101	0	3	0	0
102	0	0	2	0
103	0	0	0	3

Output Code:¶

 SELECT *
FROM
    (SELECT e.manager_id, d.name AS department, e.id AS employee_id
     FROM
     employee e
     JOIN
         department d ON e.manager_id = d.manager_id) AS SourceTable
PIVOT
    (
     COUNT(employee_id)
     FOR department IN ('HR', 'Sales', 'Engineering', 'Marketing')
    ) AS PivotTable;

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MANAGER_ID	‘HR’	‘Sales’	‘Engineering’	‘Marketing’
100	1	0	0	0
101	0	3	0	0
102	0	0	2	0
103	0	0	0	3

Unpivot Clause¶

Note

Column aliases cannot be used in the IN clause of the UNPIVOT query in Snowflake.

Input Code:¶

 CREATE TABLE count_by_color (quality VARCHAR, red INT, green INT, blue INT);

INSERT INTO count_by_color VALUES ('high', 15, 20, 7);
INSERT INTO count_by_color VALUES ('normal', 35, NULL, 40);
INSERT INTO count_by_color VALUES ('low', 10, 23, NULL);


SELECT *
FROM (SELECT red, green, blue FROM count_by_color) UNPIVOT (
    cnt FOR color IN (red, green, blue)
);

SELECT *
FROM (SELECT red, green, blue FROM count_by_color) UNPIVOT (
    cnt FOR color IN (red r, green as g, blue)
);

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COLOR	CNT
RED	15
RED	35
RED	10
GREEN	20
GREEN	23
BLUE	7
BLUE	40

Output Code:¶

 CREATE TABLE count_by_color (quality VARCHAR, red INT, green INT, blue 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 count_by_color
VALUES ('high', 15, 20, 7);
INSERT INTO count_by_color
VALUES ('normal', 35, NULL, 40);
INSERT INTO count_by_color
VALUES ('low', 10, 23, NULL);


SELECT *
FROM (SELECT red, green, blue FROM
            count_by_color
    ) UNPIVOT (
    cnt FOR color IN (red, green, blue)
);

SELECT *
FROM (SELECT red, green, blue FROM
            count_by_color
) UNPIVOT (
    cnt FOR color IN (red
                          !!!RESOLVE EWI!!! /*** SSC-EWI-RS0005 - COLUMN ALIASES CANNOT BE USED IN THE IN CLAUSE OF THE PIVOT/UNPIVOT QUERY IN SNOWFLAKE. ***/!!!
 r, green
          !!!RESOLVE EWI!!! /*** SSC-EWI-RS0005 - COLUMN ALIASES CANNOT BE USED IN THE IN CLAUSE OF THE PIVOT/UNPIVOT QUERY IN SNOWFLAKE. ***/!!!
 as g, blue)
);

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COLOR	CNT
RED	15
GREEN	20
BLUE	7
RED	35
BLUE	40
RED	10
GREEN	23

Related EWIs¶

SSC-EWI-RS0005: Column aliases cannot be used in the IN clause of the PIVOT/UNPIVOT query in Snowflake.

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 group_by_clause [, ...]

group_by_clause := {
    expr |
    GROUPING SETS ( () | group_by_clause [, ...] ) |
    ROLLUP ( expr [, ...] ) |
    CUBE ( expr [, ...] )
    }

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Sample Source Patterns

Grouping sets

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
FROM employee
GROUP BY GROUPING SETS 
    ((manager_id), ())
ORDER BY manager_id;

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MANAGER_ID	TOTAL_EMPLOYEES
100	1
101	3
102	2
103	3
104	3
null	1
null	13

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);

SELECT
    manager_id,
    COUNT(id) AS total_employees
FROM
    employee
GROUP BY GROUPING SETS
    ((manager_id), ())
ORDER BY manager_id;

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MANAGER_ID	TOTAL_EMPLOYEES
100	1
101	3
102	2
103	3
104	3
null	1
null	13

Group by Cube

Input Code:

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

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MANAGER_ID	TOTAL_EMPLOYEES
100	1
101	3
102	2
103	3
104	3
null	1
null	13

Output Code:

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

Copy

MANAGER_ID	TOTAL_EMPLOYEES
100	1
101	3
102	2
103	3
104	3
null	1
null	13

Group by Rollup

Input Code:

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

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MANAGER_ID	TOTAL_EMPLOYEES
100	1
101	3
102	2
103	3
104	3
null	1
null	13

Output Code:

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

Copy

MANAGER_ID	TOTAL_EMPLOYEES
100	1
101	3
102	2
103	3
104	3
null	1
null	13

Related EWIs

There are no known issues.

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 ]

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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);

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

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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": "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);

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

Related EWIs¶

There are no known issues.

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
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

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": "11/05/2024",  "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);

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

Related EWIs

There are no known issues.

QUALIFY clause

Description

The QUALIFY clause filters results of a previously computed window function according to user‑specified search conditions. You can use the clause to apply filtering conditions to the result of a window function without using a subquery. (Redshift SQL Language Reference QUALIFY Clause)

The QUALIFY clause is supported in Snowflake.

Grammar Syntax¶

 QUALIFY condition

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Sample Source Patterns¶

Input Code:¶

 CREATE TABLE store_sales 
(
    ss_sold_date DATE, 
    ss_sold_time TIME, 
    ss_item TEXT, 
    ss_sales_price FLOAT
);

INSERT INTO store_sales VALUES ('2022-01-01', '09:00:00', 'Product 1', 100.0),
                               ('2022-01-01', '11:00:00', 'Product 2', 500.0),
                               ('2022-01-01', '15:00:00', 'Product 3', 20.0),
                               ('2022-01-01', '17:00:00', 'Product 4', 1000.0),
                               ('2022-01-01', '18:00:00', 'Product 5', 30.0),
                               ('2022-01-02', '10:00:00', 'Product 6', 5000.0),
                               ('2022-01-02', '16:00:00', 'Product 7', 5.0);

SELECT *
FROM store_sales ss
WHERE ss_sold_time > time '12:00:00'
QUALIFY row_number()
OVER (PARTITION BY ss_sold_date ORDER BY ss_sales_price DESC) <= 2;

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SS_SOLD_DATE	SS_SOLD_TIME	SS_ITEM	SS_SALES_PRICE
2022-01-01	17:00:00	Product 4	1000
2022-01-01	18:00:00	Product 5	30
2022-01-02	16:00:00	Product 7	5

Output Code:¶

 CREATE TABLE store_sales
(
    ss_sold_date DATE,
    ss_sold_time TIME,
    ss_item TEXT,
    ss_sales_price FLOAT
)
COMMENT = '{ "origin": "sf_sc", "name": "snowconvert", "version": {  "major": 0,  "minor": 0,  "patch": "0" }, "attributes": {  "component": "redshift",  "convertedOn": "11/05/2024",  "domain": "test" }}';

INSERT INTO store_sales
VALUES ('2022-01-01', '09:00:00', 'Product 1', 100.0),
                               ('2022-01-01', '11:00:00', 'Product 2', 500.0),
                               ('2022-01-01', '15:00:00', 'Product 3', 20.0),
                               ('2022-01-01', '17:00:00', 'Product 4', 1000.0),
                               ('2022-01-01', '18:00:00', 'Product 5', 30.0),
                               ('2022-01-02', '10:00:00', 'Product 6', 5000.0),
                               ('2022-01-02', '16:00:00', 'Product 7', 5.0);

SELECT *
FROM
    store_sales ss
WHERE ss_sold_time > time '12:00:00'
QUALIFY row_number()
OVER (PARTITION BY ss_sold_date ORDER BY ss_sales_price DESC) <= 2;

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SS_SOLD_DATE	SS_SOLD_TIME	SS_ITEM	SS_SALES_PRICE
2022-01-02	16:00:00	Product 7	5
2022-01-01	17:00:00	Product 4	1000
2022-01-01	18:00:00	Product 5	30

Related EWIs¶

There are no known issues.

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 ] [, ...]

Copy

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
FROM employee;

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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": "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);

SELECT TOP 5 id, name, manager_id
FROM
    employee;

Copy

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
FROM employee;

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

Output Code:¶

 SELECT ALL manager_id
FROM
    employee;

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

DISTINCT¶

Input Code:¶

SELECT DISTINCT manager_id
FROM employee;

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MANAGER_ID
null
100
101
102
103
104

Output Code:¶

SELECT DISTINCT manager_id
FROM 
    employee;

Copy

MANAGER_ID
null
100
101
102
103
104

Related EWIs¶

There are no known issues.

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
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:

 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;

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

Related EWIs

There are no known issues.

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
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": "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);

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

Related EWIs¶

There are no known issues.

WITH clause¶

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 )

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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;

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

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

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;

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EMPLOYEE	MANAGER
Carlos	null
John	Carlos
Jorge	John
Kwaku	John
Liu	John
Mateo	Jorge
Sofía	Jorge
Nikki	Kwaku
Paulo	Kwaku
Richard	Kwaku
Saanvi	Paulo
Shirley	Paulo
Zhang	Paulo

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;

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EMPLOYEE	MANAGER
John	Carlos
Jorge	John
Kwaku	John
Liu	John
Mateo	Jorge
Sofía	Jorge
Nikki	Kwaku
Paulo	Kwaku
Richard	Kwaku
Saanvi	Paulo
Shirley	Paulo
Zhang	Paulo
Carlos	null

Related EWIs

There are no known issues.