SnowConvert: Redshift SELECT-Anweisung¶

SELECT¶

Beschreibung¶

Gibt Zeilen aus Tabellen, Ansichten und benutzerdefinierten Funktionen zurück. (Redshift SQL-Referenz: SELECT-Anweisung)

Grammatikalische 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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Weitere Informationen finden Sie unter den folgenden Links:

WITH-Klausel
SELECT-Liste
FROM-Klausel
WHERE-Klausel
CONNECT BY-Klausel
GROUP BY-Klausel
HAVING-Klausel
QUALIFY-Klausel
UNION, INTERSECT, und EXCEPT
ORDER BY-Klausel

CONNECT BY-Klausel.¶

Beschreibung¶

Die CONNECT BY-Klausel gibt die Beziehung zwischen den Zeilen in einer Hierarchie an. Sie können CONNECT BY verwenden, um Zeilen in einer hierarchischen Reihenfolge auszuwählen, indem Sie die Tabelle mit sich selbst verbinden und die hierarchischen Daten verarbeiten. (Redshift SQL-Referenz: CONNECT BY-Klausel)

Die CONNECTBY-Klausel wird in Snowflake unterstützt.

Grammar Syntax

 [START WITH start_with_conditions]
CONNECT BY connect_by_conditions

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Beispielhafte Quellcode-Muster

Eingabecode:

 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

Ausgabecode:

 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

Es gibt keine bekannten Probleme.

FROM-Klausel.

Description

Die FROM-Klausel in einer Abfrage listet die Tabellenverweise (Tabellen, Ansichten und Unterabfragen) auf, aus denen die Daten ausgewählt werden. Wenn mehrere Tabellenreferenzen aufgeführt sind, müssen die Tabellen unter Verwendung der entsprechenden Syntax entweder in der FROM-Klausel oder der WHERE-Klausel verbunden werden. Wenn keine Verknüpfungskriterien angegeben werden, verarbeitet das System die Abfrage als Cross-Join. (Redshift SQL-Referenz: FROM-Klausel)

Warnung

Die FROM-Klausel wird in Snowflake teilweise unterstützt. Entpivotieren von Objekten wird derzeit nicht unterstützt.

Grammatikalische 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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Beispielhafte Quellcode-Muster¶

Verknüpfungstypen (Join Types)¶

Snowflake unterstützt alle Arten von Verknüpfungen. Weitere Informationen finden Sie in der Dokumentation JOIN.

Eingabecode:¶

 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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Innere Verknüpfung (Inner Join)¶

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

Linke Verknüpfung (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

Rechte Verknüpfung (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

Vollständige Verknüpfung (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

Ausgabecode:¶

 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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Innere Verknüpfung (Inner Join)¶

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

Linke Verknüpfung (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

Rechte Verknüpfung (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

Vollständige Verknüpfung (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-Klausel¶

Bemerkung

Spalten-Aliase können in der IN-Klausel der PIVOT-Abfrage in Snowflake nicht verwendet werden.

Eingabecode:¶

 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

Ausgabecode:¶

 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-Klausel¶

Bemerkung

Spalten-Aliase können in der IN-Klausel der UNPIVOT-Abfrage in Snowflake nicht verwendet werden.

Eingabecode:¶

 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

Ausgabecode:¶

 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

Zugehörige EWIs¶

[SSC-EWI-RS0005](../../general/technical-documentation/issues-and-troubleshooting/conversion-issues/redshiftEWI. md#ssc-ewi-rs0005): Spalten-Aliase können in der IN-Klausel der PIVOT/UNPIVOT-Abfrage in Snowflake nicht verwendet werden.

GROUP BY-Klausel.¶

Beschreibung¶

Die GROUP BY-Klausel identifiziert die Gruppierungsspalten für die Abfrage. Gruppierungsspalten müssen deklariert werden, wenn die Abfrage Aggregate mit Standardfunktionen wie SUM, AVG und COUNT berechnet. (Redshift SQL-Referenz: GROUP BY-Klausel)

Die GROUP BY-Klausel wird in Snowflake vollständig unterstützt.

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

Gruppierungssätze

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;

Copy

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

Es gibt keine bekannten Probleme.

HAVING-Klausel.

Description

Die HAVING-Klausel wendet eine Bedingung auf das gruppierte Resultset an, das eine Abfrage zurückgibt. (Redshift SQL-Referenz: HAVING-Klausel)

Die HAVING-Klausel wird in Snowflake vollständig unterstützt.

Grammatikalische Syntax¶

 [ HAVING condition ]

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Beispielhafte Quellcode-Muster¶

Eingabecode:¶

 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;

Copy

MANAGER_ID	TOTAL_EMPLOYEES
101	3
103	3
104	3

Ausgabecode:¶

 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

Zugehörige EWIs¶

Es gibt keine bekannten Probleme.

ORDER BY-Klausel.¶

Beschreibung¶

Die ORDER BY-Klausel sortiert das Resultset einer Abfrage. (Redshift SQL-Referenz: ORDER BY-Klausel)

The ORDER BY-Klausel wird in Snowflake vollständig unterstützt.

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

Es gibt keine bekannten Probleme.

QUALIFY-Klausel.

Description

Die QUALIFY-Klausel filtert die Ergebnisse einer zuvor berechneten Fensterfunktion nach den vom Benutzer angegebenen Suchbedingungen. Sie können die Klausel verwenden, um Filterbedingungen auf das Ergebnis einer Fensterfunktion anzuwenden, ohne eine Unterabfrage zu verwenden. (Redshift SQL-Referenz: QUALIFY-Klausel)

Die QUALIFY-Klausel wird in Snowflake unterstützt.

Grammatikalische Syntax¶

 QUALIFY condition

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Beispielhafte Quellcode-Muster¶

Eingabecode:¶

 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	Produkt 4	1000
2022-01-01	18:00:00	Produkt 5	30
2022-01-02	16:00:00	Produkt 7	5

Ausgabecode:¶

 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	Produkt 7	5
2022-01-01	17:00:00	Produkt 4	1000
2022-01-01	18:00:00	Produkt 5	30

Zugehörige EWIs¶

Es gibt keine bekannten Probleme.

SELECT-Liste¶

Beschreibung¶

Die SELECT-Liste nennt die Spalten, Funktionen und Ausdrücke, die die Abfrage zurückgeben soll. Die Liste stellt die Ausgabe der Abfrage dar. (Redshift SQL-Referenz: SELECT-Liste)

Die Startoptionen für Abfragen werden in Snowflake vollständig unterstützt. Denken Sie nur daran, dass in Snowflake die Optionen DISTINCT und ALL am Anfang der Abfrage stehen müssen.

Bemerkung

Wenn Ihre Anwendung in Redshift Fremdschlüssel oder ungültige Primärschlüssel zulässt, kann dies dazu führen, dass Abfragen falsche Ergebnisse liefern. Eine SELECT DISTINCT-Abfrage könnte zum Beispiel doppelte Zeilen zurückgeben, wenn die Primärschlüsselspalte nicht alle eindeutigen Werte enthält. (Redshift SQL-Referenz: SELECT-Liste)

Grammatikalische Syntax¶

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

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Beispielhafte Quellcode-Muster¶

TOP-Klausel¶

Eingabecode:¶

 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;

Copy

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

Ausgabecode:¶

 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¶

Eingabecode:¶

SELECT ALL manager_id
FROM employee;

Copy

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

Ausgabecode:¶

 SELECT ALL manager_id
FROM
    employee;

Copy

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

DISTINCT¶

Eingabecode:¶

SELECT DISTINCT manager_id
FROM employee;

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MANAGER_ID
null
100
101
102
103
104

Ausgabecode:¶

SELECT DISTINCT manager_id
FROM 
    employee;

Copy

MANAGER_ID
null
100
101
102
103
104

Zugehörige EWIs¶

Es gibt keine bekannten Probleme.

UNION, INTERSECT und EXCEPT¶

Beschreibung¶

Die UNION-, INTERSECT- und EXCEPT-Satzperatoren werden verwendet, um die Ergebnisse von zwei separaten Abfrageausdrücken zu vergleichen und zusammenzuführen. (Redshift SQL-Referenz: Satzoperatoren)

Set-Operatoren werden in Snowflake vollständig unterstützt.

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;

Copy

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;

Copy

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

Related EWIs

Es gibt keine bekannten Probleme.

WHERE-Klausel.

Description

Die WHERE-Klausel enthält Bedingungen, die entweder Tabellen verbinden oder Prädikate auf Spalten in Tabellen anwenden. (Redshift SQL-Referenz: WHERE-Klausel)

Die WHERE-Klausel wird in Snowflake vollständig unterstützt.

Grammatikalische Syntax¶

 [ WHERE condition ]

Copy

Beispielhafte Quellcode-Muster¶

Eingabecode:¶

 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

Ausgabecode:¶

 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 '\\';

Copy

ID	NAME	MANAGER_ID
101	John	100
102	Jorge	101

Zugehörige EWIs¶

Es gibt keine bekannten Probleme.

WITH-Klausel.¶

Beschreibung¶

Eine WITH-Klausel ist eine optionale Klausel, die der SELECT-Liste in einer Abfrage vorausgeht. Die WITH-Klausel definiert eine oder mehrere common_table_expressions. Jeder gemeinsame Tabellenausdruck (CTE) definiert eine temporäre Tabelle, die einer View-Definition ähnlich ist. Sie können diese temporären Tabellen in der FROM-Klausel referenzieren. (Redshift SQL-Referenz: WITH-Klausel)

Die WITH-Klausel wird in Snowflake vollständig unterstützt.

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

Rekursive 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;

Copy

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;

Copy

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

Nicht-rekursive 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;

Copy

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;

Copy

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

Es gibt keine bekannten Probleme.