UPDATE a whole row in PL/pgSQL - postgresql

Question

I have plpgsql function:
CREATE OR REPLACE FUNCTION test() RETURNS VOID AS
$$
DECLARE
my_row my_table%ROWTYPE;
BEGIN
SELECT * INTO my_row FROM my_table WHERE id='1';
my_row.date := now();
END;
$$ LANGUAGE plpgsql;
I would like to know if it's possible to directly UPDATE my_row record.
The only way I've found to do it now is:
UPDATE my_table SET date=now() WHERE id='1';
Note this is only an example function, the real one is far more complex than this.
I'm using PostgreSQL 9.2.
UPDATE:
Sorry for the confusion, what I wanted to say is:
SELECT * INTO my_row FROM my_table INTO my_row WHERE id='1';
make_lots_of_complicated_modifications_to(my_row, other_complex_parameters);
UPDATE my_row;
I.e. Use my_row to persist information in the underlying table. I have lots of parameters to update.

Answer 1

I would like to know if it's possible to directly update "my_row"
record.
It is.
You can update columns of a row or record type in plpgsql - just like you have it. It should be working, obviously?
This would update the underlying table, of course, not the variable!
UPDATE my_table SET date=now() WHERE id='1';
You are confusing two things here ...
Answer to clarification in comment
I don't think there is syntax in PostgreSQL that can UPDATE a whole row. You can UPDATE a column list, though. Consider this demo:
Note how I use thedate instead of date as column name, date is a reserved word in every SQL standard and a type name in PostgreSQL.
CREATE TEMP TABLE my_table (id serial, thedate date);
INSERT INTO my_table(thedate) VALUES (now());
CREATE OR REPLACE FUNCTION test_up()
RETURNS void LANGUAGE plpgsql AS
$func$
DECLARE
_r my_table;
BEGIN
SELECT * INTO _r FROM my_table WHERE id = 1;
_r.thedate := now()::date + 5 ;
UPDATE my_table t
-- explicit list of columns to be to updated
SET (id, thedate) = (_r.id, _r.thedate)
WHERE t.id = 1;
END
$func$;
SELECT test_up();
SELECT * FROM my_table;
However, you can INSERT a whole row easily. Just don't supply a column list for the table (which you normally should, but in this case it is perfectly ok, not to).
As an UPDATE is internally a DELETE followed by an INSERT anyway, and a function automatically encapsulates everything in a transaction, I don't see, why you couldn't use this instead:
CREATE OR REPLACE FUNCTION x.test_ delins()
RETURNS void LANGUAGE plpgsql AS
$func$
DECLARE
_r my_table;
BEGIN
SELECT * INTO _r
FROM my_table WHERE id = 1;
_r.thedate := now()::date + 10;
DELETE FROM my_table t WHERE t.id = 1;
INSERT INTO my_table SELECT _r.*;
END
$func$;

Answer 2

I managed to get this working in PLPGSQL in a couple of lines of code.
Given a table called table in a schema called example, and a record of the same type declared as _record, you can update all the columns in the table to match the record using the following hack:
declare _record example.table;
...
-- get the columns in the correct order, as a string
select string_agg(format('%I', column_name), ',' order by ordinal_position)
into _columns
from information_schema.columns
where table_schema='example' and table_name='table';
execute 'update example.table set (' || _columns || ') = row($1.*) where pkey=$2'
using _record, _record.pkey;
In the above example, of course, _record.pkey is the table's primary key.

Answer 3

Postgresql has not set row in update.
If you wont update full row you should assign value for each column separately

Answer 4

yes, its possible to update / append the row-type variable,
CREATE OR REPLACE FUNCTION test() RETURNS VOID AS $$
DECLARE
my_row my_table%ROWTYPE;
BEGIN
SELECT * INTO my_row FROM my_table WHERE id='1';
my_row.date := now();
raise notice 'date : %; ',my_row.date;
END;
$$ LANGUAGE plpgsql;
here the raise notice will display the today's date only.
but this will not update the column date in my_table.