Re: Concatenation through SQL

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От Philippe Lang
Тема Re: Concatenation through SQL
Дата
Msg-id 6C0CF58A187DA5479245E0830AF84F4218CFBE@poweredge.attiksystem.ch
обсуждение исходный текст
Ответ на Concatenation through SQL  ("Philippe Lang" <philippe.lang@attiksystem.ch>)
Ответы Re: Concatenation through SQL
Список pgsql-sql
pgsql-sql-owner@postgresql.org wrote:
> Philippe Lang <philippe.lang@attiksystem.ch> schrieb:
>
> 2 ways:
>
> * You can use something like this:
>
> test=*# select * from a;
>  id | val
> ----+-----
>   1 | foo
>   2 | bar
> (2 rows)
>
> test=*# select array_to_string(array(select val from a), ', ');
> array_to_string -----------------
>  foo, bar
> (1 row)
>
>
> * comma-aggregate, see
> http://www.zigo.dhs.org/postgresql/#comma_aggregate

Hi,

Thanks to all who responded, in particular Andreas. I'm always amazed by
the quality of both the Postgresql database and the support in its
newsgroups.

The "comma-aggregate" worked just fine for me. Here is the final
example, for those willing to test it. This will be more than useful!

---------------------------------------
CREATE TABLE appointments
( id integer, name varchar(32), CONSTRAINT appointments_pkey PRIMARY KEY (id)
);

CREATE TABLE employees
( id integer, appointment_id integer, name varchar(32), CONSTRAINT employees_pkey PRIMARY KEY (id), CONSTRAINT
appointments_employees_fkeyFOREIGN KEY (appointment_id) 
REFERENCES appointments (id)
);

INSERT INTO appointments VALUES (1, 'app1');
INSERT INTO employees VALUES (1, 1, 'emp1');
INSERT INTO employees VALUES (2, 1, 'emp2');
INSERT INTO employees VALUES (3, 1, 'emp3');

INSERT INTO appointments VALUES (2, 'app2');
INSERT INTO employees VALUES (4, 2, 'emp1');
INSERT INTO employees VALUES (5, 2, 'emp4');

CREATE AGGREGATE array_accum (anyelement)
(  sfunc = array_append,  stype = anyarray,  initcond = '{}'
);

SELECT
appointments.name AS appointment,
(SELECT COUNT(*) FROM employees AS e where e.appointment_id =
appointments.id) AS num_employees,
(SELECT array_to_string(array_accum(name),', ') FROM employees AS e
WHERE e.appointment_id = appointments.id) AS employees
FROM appointments
---------------------------------------


Result:

---------------------------------------------------
appointment    count_employees    employees
---------------------------------------------------
app1           3                  emp1, emp2, emp3
app2           2                  emp1, emp4,
---------------------------------------------------


Merry christmas to all.

Philippe



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