Re: [GENERAL] COPY value TO STDOUT - Mailing list pgsql-general

From Pavel Stehule
Subject Re: [GENERAL] COPY value TO STDOUT
Date
Msg-id CAFj8pRDM-dBp+rCONTh9G3i42SEo6cj2z19aJQUqP0Xmwh9usw@mail.gmail.com
Whole thread Raw
In response to [GENERAL] COPY value TO STDOUT  (Denisa Cirstescu <denisa.cirstescu@asentinel.com>)
List pgsql-general
Hi

2017-01-13 16:45 GMT+01:00 Denisa Cirstescu <denisa.cirstescu@asentinel.com>:

I am not sure if this is the correct mailing list or if this is how you submit a question, but I am going to give it a try.

 

I want to COPY a value to STDOUT from PL/pgSQL language.

 

I saw that the STDOUT is not accessible from PL/pgSQL, but it is from SQL.

This is why I am trying to create an auxiliary function declared as language SQL and call that function from my PL/pgSQL code.

 

This is an example of PL/pgSQLcode:

 

DO

$do$

BEGIN

               perform printToStdout('12');

               perform printToStdout('34');

               perform printToStdout('56');

END

$do$

 

And this is the definition of the auxiliary function:

 

create or replace function printToStdout(abc text) returns void as $$

               copy (SELECT abc) to stdout;

$$ language sql;

 

 

However, this is not working, because COPY doesn’t know how to use the value of my variable named abc and it returns the error:

ERROR:  column "abc" does not exist

If I use a constant, instead of the abc variable everything works fine; the constant is printed to STDOUT.

 

Is there a way to achieve this without using an auxiliary table?

The below code does the job, but is not ok for me because of the auxiliary table that might cause performance problems in a concurrent environment with a lot of requests:

 

create table if not exists printToStdoutTable(abc text);

 

create or replace function printToStdout(abc text) returns void as $$

               delete from printToStdoutTable;

               insert into printToStdoutTable  values(abc);

               copy (SELECT * from printToStdoutTable) to stdout;

$$ language sql;



You cannot do it in plain text language. 

The identifier of column or table have not be a variable ever.

You can use dynamic SQL in PLpgSQL - where SQL command is created in run-time and there you can do what you want. 


Regards

Pavel Stehule

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