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List:       postgresql-general
Subject:    Re: dynamically generate path to output file
From:       Osvaldo Kussama <osvaldo.kussama () gmail ! com>
Date:       2022-03-26 12:48:08
Message-ID: CADgbbqjOmphi4NYkcrSPNB-at6mB=77J+AB6czUhoOGYhRU7Kw () mail ! gmail ! com
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2022-03-26 6:35 GMT-03:00, robert@redo2oo.ch <robert@redo2oo.ch>:
> 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
>
>

Look at EXECUTE statement:
43.5.4. Executing Dynamic Commands
https://www.postgresql.org/docs/current/plpgsql-statements.html#PLPGSQL-STATEMENTS-EXECUTING-DYN

Osvaldo


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