Postgresql

Question

我有一个 employer 和一个 employee table。我有一个连接 table 加入他们。

CREATE TABLE employer (id int primary key, name text);
CREATE TABLE employee (id int primary key, name text);
CREATE TABLE employer_employee_join(
    employer_id int REFERENCES employer(id),
    employee_id int REFERENCES employee(id)
);

INSERT INTO employer (id, name) VALUES (1, 'the boss');
INSERT INTO employee (id, name) VALUES (1, 'employee1');
INSERT INTO employee (id, name) VALUES (2, 'employee2');

INSERT INTO employer_employee_join (employer_id, employee_id) VALUES(1, 1);
INSERT INTO employer_employee_join (employer_id, employee_id) VALUES(1, 2);

我的 sql 查询 returns employer 并聚合 employee return 记录数组 (record[])。

SELECT 
employer.id, employer.name, array_agg((employee.id, employee.name))
FROM employer
LEFT JOIN employer_employee_join
ON employer_employee_join.employer_id = employer.id
LEFT JOIN employee
ON employee.id = employer_employee_join.employee_id
GROUP BY employer.id;

这很好用。

但是当我把它放在 PL/PGSQL 函数中时它失败了：

CREATE OR REPLACE FUNCTION _test()
    RETURNS table(id integer, name text, agg record[]) 
    LANGUAGE 'plpgsql'
AS $BODY$
begin
    SELECT 
    employer.id, employer.name, array_agg((employee.id, employee.name))
    FROM employer
    LEFT JOIN employer_employee_join
    ON employer_employee_join.employer_id = employer.id
    LEFT JOIN employee
    ON employee.id = employer_employee_join.employee_id
    GROUP BY employer.id;
end;
$BODY$

错误是

ERROR:  PL/pgSQL functions cannot accept type record[]
SQL state: 0A000

如何获取 plpgsql 函数以 return 记录数组？

（我真的不想使用 json_agg() 因为系统中的另一层在 postgresql 和我的控制之外）

Answer 1

感谢@Bergi。这是我需要的composite type。

create type employee_agg as (id int, name text);

CREATE OR REPLACE FUNCTION _test()
    RETURNS table(id integer, name text, agg employee_agg[]) 
    LANGUAGE plpgsql
AS $BODY$
begin
    return query
    SELECT 
    employer.id, 
    employer.name, 
    array_agg(row(employee.id, employee.name)::employee_agg)
    FROM employer
    LEFT JOIN employer_employee_join
    ON employer_employee_join.employer_id = employer.id
    LEFT JOIN employee
    ON employee.id = employer_employee_join.employee_id
    GROUP BY employer.id;
end;
$BODY$;

Answer 2

使用VIEW

就这样做...

CREATE VIEW _test
AS
    SELECT 
    employer.id, 
    employer.name, 
    array_agg(row(employee.id, employee.name)::employee_agg)
    FROM employer
    LEFT JOIN employer_employee_join
    ON employer_employee_join.employer_id = employer.id
    LEFT JOIN employee
    ON employee.id = employer_employee_join.employee_id
    GROUP BY employer.id;

你的函数没有参数，它只是一个查询。这是一个更好的主意。

或者，您可以使用 SQL 函数（但视图更好）。

Postgresql - return 来自 plpgsql 函数的记录[]

Postgresql - return a record[] from a plpgsql function

plpgsql

使用VIEW