Re: Concatenation through SQL
От | Philippe Lang |
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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
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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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