I am creating a database for the first time using Postgres 9.3 on MacOSX.
Let’s say I have table A
and B
. A
starts off as empty and B
as filled. I would like the number of entries in column all_names
in table B
to equal the number
for each names
in table A
like table B
below. Thus names
should contain each unique entry from all_names
and number
its count. I am not used to the syntax, yet, so I do not really know how to go about it. The birthday
column is redundant.
Table A
names | number ------+-------- Carl | 3 Bill | 4 Jen | 2
Table B
all_names | birthday -----------+------------ Carl | 17/03/1980 Carl | 22/08/1994 Carl | 04/09/1951 Bill | 02/12/2003 Bill | 11/03/1975 Bill | 04/06/1986 Bill | 08/07/2005 Jen | 05/03/2009 Jen | 01/04/1945
Would this be the correct way to go about it?
insert into a (names, number) select b.all_names, count(b.all_names) from b group by b.all_names;
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Answer
Answer to original question
Postgres allows set-returning functions (SRF) to multiply rows. generate_series()
is your friend:
INSERT INTO b (all_names, birthday) SELECT names, current_date -- AS birthday ?? FROM (SELECT names, generate_series(1, number) FROM a);
Since the introduction of LATERAL
in Postgres 9.3 you can do stick to standard SQL: the SRF moves from the SELECT
to the FROM
list:
INSERT INTO b (all_names, birthday) SELECT a.names, current_date -- AS birthday ?? FROM a, generate_series(1, a.number) AS rn
LATERAL
is implicit here, as explained in the manual:
LATERAL
can also precede a function-callFROM
item, but in this case it is a noise word, because the function expression can refer to earlier FROM items in any case.
Reverse operation
The above is the reverse operation (approximately) of a simple aggregate count()
:
INSERT INTO a (name, number) SELECT all_names, count(*) FROM b GROUP BY 1;
… which fits your updated question.
Note a subtle difference between count(*)
and count(all_names)
. The former counts all rows, no matter what, while the latter only counts rows where all_names IS NOT NULL
. If your column all_names
is defined as NOT NULL
, both return the same, but count(*)
is a bit shorter and faster.
About GROUP BY 1
: