Código:
pruebas=# select * from candidatos
pruebas-# ;
id | f_alta | f_entrevista | entrevistado
----+------------+--------------+--------------
1 | 2009-02-07 | 2000-01-01 | f
2 | 2009-02-08 | 2000-01-01 | f
3 | 2009-02-09 | 2000-01-01 | f
4 | 2009-02-10 | 2000-01-01 | f
5 | 2009-02-11 | 2000-01-01 | f
6 | 2009-02-12 | 2000-01-01 | f
7 | 2009-02-13 | 2000-01-01 | f
8 | 2009-02-14 | 2000-01-01 | f
9 | 2009-02-15 | 2000-01-01 | f
10 | 2009-02-16 | 2000-01-01 | f
11 | 2000-01-01 | 2009-02-07 | t
12 | 2000-01-01 | 2009-02-08 | t
13 | 2000-01-01 | 2009-02-09 | t
14 | 2000-01-01 | 2009-02-10 | t
15 | 2000-01-01 | 2009-02-11 | t
16 | 2000-01-01 | 2009-02-12 | t
17 | 2000-01-01 | 2009-02-13 | t
18 | 2000-01-01 | 2009-02-14 | t
19 | 2000-01-01 | 2009-02-15 | t
20 | 2000-01-01 | 2009-02-16 | t
(20 rows)
pruebas=# \d candidatos
Table "public.candidatos"
Column | Type | Modifiers
--------------+---------+---------------------------------------------------------
id | integer | not null default nextval('candidatos_id_seq'::regclass)
f_alta | date |
f_entrevista | date |
entrevistado | boolean |
Indexes:
"candidatos_pkey" PRIMARY KEY, btree (id)
pruebas=# \d candidatos_historial
Table "public.candidatos_historial"
Column | Type | Modifiers
--------------+---------+-----------
id | integer |
f_alta | date |
f_entrevista | date |
entrevistado | boolean |
col_new | integer |
pruebas=# insert into candidatos_historial select *, 1 from candidatos where id = 10;
INSERT 0 1
pruebas=# select * from candidatos_historial ;
id | f_alta | f_entrevista | entrevistado | col_new
----+------------+--------------+--------------+---------
10 | 2009-02-16 | 2000-01-01 | f | 1
Un ejemplo vale más que mil explicaciones
Salu2