Logo Questions Linux Laravel Mysql Ubuntu Git Menu
 

Use function variable in dynamic COPY statement

According to docs of PostgreSQL it is possible to copy data to csv file right from a query without using an intermediate table. I am curious how to do that.

CREATE OR REPLACE FUNCTION m_tbl(my_var integer)
    RETURNS void AS
$BODY$
DECLARE
BEGIN

   COPY (
       select my_var
   )
   TO 'c:/temp/out.csv';

END;
$$ LANGUAGE plpgsql;

I get an error: no such column 'my_var'.

like image 780
simar Avatar asked Jan 22 '26 15:01

simar


1 Answers

Yes, it is possible to COPY from any query, whether or not it refers to a table.

However, COPY is a non-plannable statement, a utility statement. It doesn't support query parameters - and query parameters are how PL/PgSQL implements the insertion of variables into statements.

So you can't use PL/PgSQL variables with COPY.

You must instead use dynamic SQL with EXECUTE. See the Pl/PgSQL documentation for examples. There are lots of examples here on Stack Overflow and on https://dba.stackexchange.com/ too.

Something like:

EXECUTE format('
   COPY (
       select %L
   )
   TO ''c:/temp/out.csv'';
', my_var);

The same applies if you want the file path to be dynamic - you'd use:

EXECUTE format('
   COPY (
       select %L
   )
   TO %L;
', my_var, 'file_name.csv');

It also works for dynamic column names but you would use %I (for identifier, like "my_name") instead of %L for literal like 'my_value'. For details on %I and %L, see the documentation for format.

like image 199
Craig Ringer Avatar answered Jan 24 '26 12:01

Craig Ringer



Donate For Us

If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!