[EMAIL PROTECTED] wrote:
> Philippe Lang <[EMAIL PROTECTED]> 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_fkey FOREIGN 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


---------------------------(end of broadcast)---------------------------
TIP 9: In versions below 8.0, the planner will ignore your desire to
       choose an index scan if your joining column's datatypes do not
       match

Reply via email to