SnowConvert:Redshift SELECT 语句¶
SELECT¶
描述¶
从表、视图和用户定义的函数中返回行。(Redshift SQL 语言参考“SELECT 语句” (https://docs.aws.amazon.com/redshift/latest/dg/r_SELECT_synopsis.html))
语法¶
[ 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 ]
有关更多信息,请参阅以下每个链接:
CONNECT BY 子句¶
描述¶
CONNECT BY
子句指定了层次结构中的行之间的关系。通过将表与其自身联接并处理分层数据,您可以使用 CONNECT BY
按分层顺序选择行。(Redshift SQL 语言参考“CONNECT BY 子句” (https://docs.aws.amazon.com/redshift/latest/dg/r_CONNECT_BY_clause.html))
Snowflake 支持 CONNECT BY 子句。
Grammar Syntax
[START WITH start_with_conditions]
CONNECT BY connect_by_conditions
示例源模式
输入代码:
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';
COUNT(*) |
---|
12 |
输出代码:
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');
COUNT(*) |
---|
12 |
相关的 EWIs
没有已知问题。
FROM 子句
Description
查询中的 FROM
子句列出了从中选择数据的表引用(表、视图和子查询)。如果列出了多个表引用,则必须在 FROM
子句或 WHERE
子句中使用相应语法联接这些表。如果未指定联接条件,则系统会将查询作为交叉联接进行处理。(Redshift SQL 语言参考“FROM 子句”) (https://docs.aws.amazon.com/redshift/latest/dg/r_FROM_clause30.html))
警告
Snowflake 部分支持 FROM 子句。目前不支持 Object 取消透视 (https://docs.aws.amazon.com/redshift/latest/dg/query-super.html#unpivoting)。
语法¶
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 ]
示例源模式¶
联接类型¶
Snowflake 支持所有类型的连接。有关更多信息,请参阅 JOIN 文档。
输入代码:¶
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;
内部联接¶
EMPLOYEE_NAME | DEPARTMENT_NAME |
---|---|
John | HR |
Jorge | Sales |
Kwaku | Sales |
Liu | Sales |
Mateo | Engineering |
Nikki | Marketing |
Paulo | Marketing |
Richard | Marketing |
Sofía | Engineering |
左联接¶
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 |
右联接¶
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 |
全连接¶
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 |
输出代码:¶
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;
内部联接¶
EMPLOYEE_NAME |
DEPARTMENT_NAME |
---|---|
John |
HR |
Jorge |
Sales |
Kwaku |
Sales |
Liu |
Sales |
Mateo |
Engineering |
Nikki |
Marketing |
Paulo |
Marketing |
Richard |
Marketing |
Sofía |
Engineering |
左联接¶
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 |
右联接¶
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 |
全连接¶
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 子句¶
备注
在 Snowflake 中,PIVOT 查询的 IN 子句中不能使用列别名。
输入代码:¶
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;
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 |
输出代码:¶
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;
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 子句¶
备注
在 Snowflake 中,UNPIVOT 查询的 IN 子句中不能使用列别名。
输入代码:¶
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)
);
COLOR |
CNT |
---|---|
RED |
15 |
RED |
35 |
RED |
10 |
GREEN |
20 |
GREEN |
23 |
BLUE |
7 |
BLUE |
40 |
输出代码:¶
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)
);
COLOR |
CNT |
---|---|
RED |
15 |
GREEN |
20 |
BLUE |
7 |
RED |
35 |
BLUE |
40 |
RED |
10 |
GREEN |
23 |
相关的 EWIs¶
SSC-EWI-RS0005: 在 Snowflake 中,PIVOT/UNPIVOT 查询的 IN 子句中不能使用列别名。
GROUP BY 子句¶
描述¶
GROUP BY
子句标识了查询的分组列。当查询使用标准函数(如 SUM
、AVG
和 COUNT
)计算聚合时,必须声明分组列。(Redshift SQL 语言参考“GROUP BY 子句” (https://docs.aws.amazon.com/redshift/latest/dg/r_GROUP_BY_clause.html))
Snowflake 完全支持 GROUP BY 子句。
Grammar Syntax
GROUP BY group_by_clause [, ...]
group_by_clause := {
expr |
GROUPING SETS ( () | group_by_clause [, ...] ) |
ROLLUP ( expr [, ...] ) |
CUBE ( expr [, ...] )
}
Sample Source Patterns
分组集
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;
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;
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;
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;
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;
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;
MANAGER_ID |
TOTAL_EMPLOYEES |
---|---|
100 |
1 |
101 |
3 |
102 |
2 |
103 |
3 |
104 |
3 |
null |
1 |
null |
13 |
Related EWIs
没有已知问题。
HAVING 子句
Description
HAVING
子句可将条件应用于查询返回的中间分组结果集。(Redshift SQL 语言参考“HAVING 子句” (https://docs.aws.amazon.com/redshift/latest/dg/r_HAVING_clause.html))
Snowflake 完全支持 HAVING 子句。
语法¶
[ HAVING condition ]
示例源模式¶
输入代码:¶
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;
MANAGER_ID |
TOTAL_EMPLOYEES |
---|---|
101 |
3 |
103 |
3 |
104 |
3 |
输出代码:¶
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;
MANAGER_ID |
TOTAL_EMPLOYEES |
---|---|
101 |
3 |
103 |
3 |
104 |
3 |
相关的 EWIs¶
没有已知问题。
ORDER BY 子句¶
描述¶
ORDER BY
子句对查询的结果集进行排序。(Redshift SQL 语言参考“Order By 子句” (https://docs.aws.amazon.com/redshift/latest/dg/r_ORDER_BY_clause.html))
Snowflake 完全支持 ORDER BY 子句。
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
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": "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;
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
没有已知问题。
QUALIFY 子句
Description
QUALIFY
子句根据用户指定的搜索条件筛选先前计算的窗口函数的结果。您可以使用该子句对窗口函数的结果应用筛选条件,而无需使用子查询。(Redshift SQL 语言参考“QUALIFY 子句” (https://docs.aws.amazon.com/redshift/latest/dg/r_QUALIFY_clause.html))
Snowflake 支持 QUALIFY 子句。
语法¶
QUALIFY condition
示例源模式¶
输入代码:¶
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;
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 |
输出代码:¶
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;
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 |
相关的 EWIs¶
没有已知问题。
SELECT 列表¶
描述¶
SELECT 列表会命名您希望查询返回的列、函数和表达式。该列表表示查询的输出。(Redshift SQL 语言参考“SELECT 列表” (https://docs.aws.amazon.com/redshift/latest/dg/r_SELECT_list.html))
Snowflake 完全支持 查询起始选项。请记住,在 Snowflake 中,DISTINCT
和 ALL
选项必须放在查询的开头。
备注
在 Redshift 中,如果应用程序允许使用外键或无效的主键,则可能导致查询返回不正确的结果。例如,如果“主键”列中不包含所有唯一值,则 SELECT DISTINCT 查询可能会返回重复行。(Redshift SQL 语言参考“SELECT 列表” (https://docs.aws.amazon.com/redshift/latest/dg/r_SELECT_list.html))
语法¶
SELECT
[ TOP number ]
[ ALL | DISTINCT ] * | expression [ AS column_alias ] [, ...]
示例源模式¶
Top 子句¶
输入代码:¶
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;
ID |
NAME |
MANAGER_ID |
---|---|---|
100 |
Carlos |
null |
101 |
John |
100 |
102 |
Jorge |
101 |
103 |
Kwaku |
101 |
110 |
Liu |
101 |
输出代码:¶
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;
ID |
NAME |
MANAGER_ID |
---|---|---|
100 |
Carlos |
null |
101 |
John |
100 |
102 |
Jorge |
101 |
103 |
Kwaku |
101 |
110 |
Liu |
101 |
ALL¶
输入代码:¶
SELECT ALL manager_id
FROM employee;
MANAGER_ID |
---|
null |
100 |
101 |
101 |
101 |
102 |
103 |
103 |
103 |
104 |
104 |
102 |
104 |
输出代码:¶
SELECT ALL manager_id
FROM
employee;
MANAGER_ID |
---|
null |
100 |
101 |
101 |
101 |
102 |
103 |
103 |
103 |
104 |
104 |
102 |
104 |
DISTINCT¶
输入代码:¶
SELECT DISTINCT manager_id
FROM employee;
MANAGER_ID |
---|
null |
100 |
101 |
102 |
103 |
104 |
输出代码:¶
SELECT DISTINCT manager_id
FROM
employee;
MANAGER_ID |
---|
null |
100 |
101 |
102 |
103 |
104 |
相关的 EWIs¶
没有已知问题。
UNION、INTERSECT 和 EXCEPT¶
描述¶
UNION
、INTERSECT
和 EXCEPT
集合运算符 用于比较和合并两个单独的查询表达式的结果。(Redshift SQL 语言参考“集合运算符” (https://docs.aws.amazon.com/redshift/latest/dg/r_UNION.html))
Snowflake 完全支持 集合运算符。
Grammar Syntax
query
{ UNION [ ALL ] | INTERSECT | EXCEPT | MINUS }
query
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;
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;
ID |
NAME |
MANAGER_ID |
---|---|---|
102 |
Jorge |
101 |
103 |
Kwaku |
101 |
110 |
Liu |
101 |
106 |
Mateo |
102 |
201 |
Sofía |
102 |
Related EWIs
没有已知问题。
WHERE 子句
Description
WHERE
子句中包含联接表或将谓词应用于表列的条件。(Redshift SQL 语言参考“WHERE 子句” (https://docs.aws.amazon.com/redshift/latest/dg/r_WHERE_clause.html))
Snowflake 完全支持 WHERE 子句。
语法¶
[ WHERE condition ]
示例源模式¶
输入代码:¶
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%';
ID |
NAME |
MANAGER_ID |
---|---|---|
101 |
John |
100 |
102 |
Jorge |
101 |
输出代码:¶
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 '\\';
ID |
NAME |
MANAGER_ID |
---|---|---|
101 |
John |
100 |
102 |
Jorge |
101 |
相关的 EWIs¶
没有已知问题。
WITH 子句¶
描述¶
WITH
子句是查询中的可选子句,需置于 SELECT 列表之前。WITH
子句定义了一个或多个 common_table_expressions。每个公用表表达式 (CTE) 都定义了一个临时表,这类似于视图定义。您可以在 FROM
子句中引用这些临时表。(Redshift SQL 语言参考“WITH 子句” (https://docs.aws.amazon.com/redshift/latest/dg/r_WITH_clause.html))
Snowflake 完全支持 WITH 子句。
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
递归形式
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;
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;
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 |
非递归形式
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;
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;
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
没有已知问题。