SnowConvert: Redshift SELECT INTO 문¶
SELECT INTO¶
설명¶
테이블, 뷰 및 사용자 정의 함수에서 행을 반환하여 새 테이블에 삽입합니다. (Redshift SQL Language Reference SELECT 문)
문법 구문¶
[ 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 ]
자세한 내용은 다음 각 링크를 참조하십시오.
FROM 절¶
설명¶
쿼리의 FROM 절은 데이터가 선택된 테이블 참조(테이블, 뷰 및 하위 쿼리)를 목록으로 표시합니다. 여러 테이블 참조가 목록에 있는 경우 FROM 절 또는 WHERE 절에서 적절한 구문을 사용하여 테이블을 조인해야 합니다. 조인 조건이 지정되지 않으면 시스템은 쿼리를 교차 조인으로 처리합니다. (Redshift SQL Language Reference FROM 절)
경고
FROM 절은 Snowflake에서 부분적으로 지원됩니다. Object unpivoting 은 현재 지원되지 않습니다.
문법 구문¶
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 ]
샘플 소스 패턴¶
입력 코드:¶
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 |
마케팅 |
Paulo |
마케팅 |
Richard |
마케팅 |
Sofía |
Engineering |
출력 코드:¶
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;
| EMPLOYEE_NAME | DEPARTMENT_NAME |
|---|---|
| John | HR |
| Jorge | Sales |
| Kwaku | Sales |
| Liu | Sales |
| Mateo | Engineering |
| Nikki | Marketing |
| Paulo | Marketing |
| Richard | Marketing |
| Sofía | Engineering |
Known Issues¶
알려진 문제는 없습니다.
GROUP BY 절¶
설명¶
GROUP BY 절은 쿼리의 그룹화 열을 식별합니다. 그룹화 열은 쿼리가 SUM, AVG, COUNT 같은 표준 함수로 집계를 계산할 때 선언해야 합니다. (Redshift SQL Language Reference GROUP BY 절)
GROUP BY 절은 Snowflake에서 완전히 지원됩니다.
Grammar Syntax
GROUP BY expression [, ...]
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;
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;
MANAGER_ID |
TOTAL_EMPLOYEES |
|---|---|
100 |
1 |
101 |
3 |
102 |
2 |
103 |
3 |
104 |
3 |
1 |
Known Issues
알려진 문제는 없습니다.
Related EWIs.
관련 EWIs 가 없습니다.
HAVING 절
Description
HAVING 절은 쿼리가 반환하는 중간 그룹화된 결과 세트에 조건을 적용합니다. (Redshift SQL Language Reference HAVING 절)
HAVING 절 은 Snowflake에서 완전히 지원됩니다.
문법 구문¶
[ HAVING condition ]
샘플 소스 패턴¶
입력 코드:¶
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;
MANAGER_ID |
TOTAL_EMPLOYEES |
|---|---|
101 |
3 |
103 |
3 |
104 |
3 |
출력 코드:¶
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;
MANAGER_ID |
TOTAL_EMPLOYEES |
|---|---|
101 |
3 |
103 |
3 |
104 |
3 |
Known Issues¶
알려진 문제는 없습니다.
관련 EWIs.¶
관련 EWIs 가 없습니다.
LIMIT 및 OFFSET 절¶
설명¶
LIMIT 및 OFFSET 절은 숫자에 지정된 행 수를 검색하고 건너뜁니다.
LIMIT 및 OFFSET 절은 Snowflake에서 완전히 지원됩니다.
Grammar Syntax
[ LIMIT { number | ALL } ]
[ OFFSET start ]
Sample Source Patterns
LIMIT 숫자
Input Code:
SELECT id, name, manager_id, salary
INTO limited_employees
FROM employee
LIMIT 5;
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;
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;
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;
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 |
LIMIT을 제외한 OFFSET
Snowflake는 LIMIT 가 없는 OFFSET 을 지원하지 않습니다. LIMIT 는 기본값인 NULL 로 변환한 후 추가되며, 이 값이 기본 LIMIT 입니다.
Input Code:
SELECT id, name, manager_id, salary
INTO limited_employees
FROM employee
OFFSET 5;
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;
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
알려진 문제는 없습니다.
Related EWIs.
관련 EWIs 가 없습니다.
로컬 변수 및 매개 변수
Description
Redshift는 또한 문이 저장 프로시저 내에서 실행될 때 SELECT INTO 변수를 허용합니다.
이 패턴은 Snowflake에서 완벽하게 지원됩니다.
문법 구문¶
SELECT [ select_expressions ] INTO target [ select_expressions ] FROM ...;
샘플 소스 패턴¶
왼쪽에 식이 있는 SELECT INTO¶
입력 코드:¶
CREATE OR REPLACE PROCEDURE test_sp1(out param1 int)
AS $$
DECLARE
var1 int;
BEGIN
select 10, 100 into param1, var1;
END;
$$ LANGUAGE plpgsql;
param1 |
|---|
10 |
출력 코드:¶
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 $$
DECLARE
var1 int;
BEGIN
select 10, 100 into
: param1,
: var1;
RETURN OBJECT_CONSTRUCT('param1', :param1);
END;
$$;
TEST_SP1 |
|---|
{ “param1”: 10 } |
오른쪽에 식이 있는 SELECT INTO¶
입력 코드:¶
CREATE OR REPLACE PROCEDURE test_sp1(out param1 int)
AS $$
DECLARE
var1 int;
BEGIN
select into param1, var1 10, 100;
END;
$$ LANGUAGE plpgsql;
param1 |
|---|
10 |
출력 코드:¶
Snowflake는 SELECT INTO 에 대해 이 문법을 지원하지 않으므로 식은 INTO 의 왼쪽으로 이동합니다.
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 $$
DECLARE
var1 int;
BEGIN
select
10, 100
into
: param1,
: var1;
RETURN OBJECT_CONSTRUCT('param1', :param1);
END;
$$;
TEST_SP1 |
|---|
{ “param1”: 10 } |
Known Issues¶
알려진 문제는 없습니다.
관련 EWIs.¶
관련 EWIs 가 없습니다.
ORDER BY 절¶
설명¶
ORDER BY 절은 쿼리 결과 세트를 정렬합니다. (Redshift SQL Language Reference Order By 절)
ORDER BY 절 은 Snowflake에서 완전히 지원됩니다.
Grammar Syntax
[ ORDER BY expression [ ASC | DESC ] ]
[ NULLS FIRST | NULLS LAST ]
[ LIMIT { count | ALL } ]
[ OFFSET start ]
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;
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;
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
알려진 문제는 없습니다.
Related EWIs.
관련 EWIs 가 없습니다.
SELECT 목록
Description
SELECT 목록은 쿼리에서 반환할 열, 함수 및 식의 이름을 지정합니다. 목록은 쿼리의 출력을 나타냅니다. (Redshift SQL Language Reference SELECT 목록)
쿼리 시작 옵션 은 Snowflake에서 완벽하게 지원됩니다. Snowflake에서는 DISTINCT 및 ALL 옵션이 쿼리 시작 부분에 있어야 한다는 점을 기억하십시오.
참고
애플리케이션에서 외래 키 또는 유효하지 않은 기본 키를 허용하는 경우 Redshift에서 쿼리 결과가 잘못 반환될 수 있습니다. 예를 들어, 기본 키 열에 모든 고유 값이 포함되어 있지 않은 경우 SELECT DISTINCT 쿼리는 중복 행을 반환할 수 있습니다. (Redshift SQL Language Reference SELECT 목록)
문법 구문¶
SELECT
[ TOP number ]
[ ALL | DISTINCT ] * | expression [ AS column_alias ] [, ...]
샘플 소스 패턴¶
Top 절¶
입력 코드:¶
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;
ID |
NAME |
MANAGER_ID |
|---|---|---|
100 |
Carlos |
Null |
101 |
John |
100 |
102 |
Jorge |
101 |
103 |
Kwaku |
101 |
110 |
Liu |
101 |
출력 코드:¶
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;
ID |
NAME |
MANAGER_ID |
|---|---|---|
100 |
Carlos |
Null |
101 |
John |
100 |
102 |
Jorge |
101 |
103 |
Kwaku |
101 |
110 |
Liu |
101 |
ALL¶
입력 코드:¶
SELECT ALL manager_id
INTO manager
FROM employee;
MANAGER_ID |
|---|
Null |
100 |
101 |
101 |
101 |
102 |
103 |
103 |
103 |
104 |
104 |
102 |
104 |
출력 코드:¶
CREATE TABLE IF NOT EXISTS manager AS
SELECT ALL manager_id
FROM
employee;
MANAGER_ID |
|---|
Null |
100 |
101 |
101 |
101 |
102 |
103 |
103 |
103 |
104 |
104 |
102 |
104 |
DISTINCT¶
입력 코드:¶
SELECT DISTINCT manager_id
INTO manager
FROM employee;
MANAGER_ID |
|---|
Null |
100 |
101 |
102 |
103 |
104 |
출력 코드:¶
CREATE TABLE IF NOT EXISTS manager AS
SELECT DISTINCT manager_id
FROM
employee;
MANAGER_ID |
|---|
Null |
100 |
101 |
102 |
103 |
104 |
Known Issues¶
알려진 문제는 없습니다.
관련 EWIs.¶
관련 EWIs 가 없습니다.
UNION, INTERSECT, EXCEPT¶
설명¶
UNION, INTERSECT, EXCEPT _set 연산자_는 2개의 개별 쿼리 식의 결과를 비교하고 병합하는 데 사용됩니다. (Redshift SQL Language Reference Set 연산자)
Set 연산자 는 Snowflake에서 완벽하게 지원됩니다.
Grammar Syntax
query
{ UNION [ ALL ] | INTERSECT | EXCEPT | MINUS }
query
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;
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;
ID |
NAME |
MANAGER_ID |
|---|---|---|
102 |
Jorge |
101 |
103 |
Kwaku |
101 |
110 |
Liu |
101 |
106 |
Mateo |
102 |
201 |
Sofía |
102 |
Known Issues
알려진 문제는 없습니다.
Related EWIs.
관련 EWIs 가 없습니다.
WHERE 절
Description
WHERE 절에는 테이블을 조인하거나 테이블의 열에 조건자를 적용하는 조건이 포함되어 있습니다. (Redshift SQL Language Reference WHERE 절)
WHERE 절은 Snowflake에서 완전히 지원됩니다.
문법 구문¶
[ WHERE condition ]
샘플 소스 패턴¶
입력 코드:¶
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%';
ID |
NAME |
MANAGER_ID |
|---|---|---|
101 |
John |
100 |
102 |
Jorge |
101 |
출력 코드:¶
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 '\\';
ID |
NAME |
MANAGER_ID |
|---|---|---|
101 |
John |
100 |
102 |
Jorge |
101 |
Known Issues¶
알려진 문제는 없습니다.
관련 EWIs.¶
관련 EWIs 가 없습니다.
WITH 절¶
설명¶
WITH 절은 쿼리에서 SELECT INTO 앞에 오는 선택적 절입니다. WITH 절은 1개 이상의 _common_table_expression_을 정의합니다. 각 공통 테이블 식(CTE)은 뷰 정의와 유사한 임시 테이블을 정의합니다. FROM 절에서 이러한 임시 테이블을 참조할 수 있습니다. (Redshift SQL Language Reference WITH 절)
WITH 절 은 Snowflake에서 완전히 지원됩니다.
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
비재귀 형식
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;
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;
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 |
재귀 형식
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;
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;
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
알려진 문제는 없습니다.
Related EWIs.
관련 EWIs 가 없습니다.