dynamically generate path to output file

From: "robert(at)redo2oo(dot)ch" <robert(at)redo2oo(dot)ch>
To: pgsql-general(at)lists(dot)postgresql(dot)org, robert rottermann <robert(at)redcor(dot)ch>
Subject: dynamically generate path to output file
Date: 2022-03-26 09:35:23
Message-ID: bd57256b-162c-ac18-ecc0-968ee582b744@redo2oo.ch
Views: Raw Message | Whole Thread | Download mbox | Resend email
Thread:
Lists: pgsql-general

Hi Friends

I would like to generate the path of a CSV file to which I output a query.

This is what I try:

CREATEORREPLACEPROCEDUREexport_cvs(
home_dir varchar
)
AS
$BODY$
DECLARE
OUTFILE varchar;
BEGIN
OUTFILE = (home_dir || '/tmp/company.csv');
copy(
select'company_'||id as"External ID",
nameas"Name",'True'as"Is a Company",
email,
phone ,
company_registry
fromres_company
) TOOUTFILE withCSV HEADER;
END;
$BODY$
LANGUAGEplpgsql;

this produces a syntax error:

psql:export_contacts_short.sql:21: ERROR:  syntax error at or near "OUTFILE"
LINE 17:     ) TO OUTFILE  with CSV HEADER;

It works fine when I replace "TO OUTFILE" with a hard coded string.

Can anybody of you give me a hint how to do that?

Thanks a lot.

Robert

Responses

Browse pgsql-general by date

  From Date Subject
Next Message Shaozhong SHI 2022-03-26 12:12:05 In what situation, a line feature could not be visible in PgAdmin?
Previous Message Laurenz Albe 2022-03-26 05:08:40 Re: Leading comments and client applications