I got to know that I can use row_to_json to return json output
For example If my query is:
select * from sample;
I can rewrite it as follows to return json output:
select row_to_json(sample) from sample;
But one thing I am trying to achieve is the same functionality in the function.
To give you an example, here is the function returning table:
CREATE FUNCTION find_val(val text)
RETURNS SETOF sample AS
$$
BEGIN
RETURN QUERY
SELECT * FROM sample where $1 = ANY(col4);
END;
$$
LANGUAGE 'plpgsql';
Now instead of rows, I want to return JSON output from my function. How can I do that ?
Here is what I have tried so far:
native=> CREATE FUNCTION find_val(val text)
RETURNS SETOF sample AS
$$
BEGIN
RETURN QUERY
SELECT row_to_json(sample) FROM sample where $1 = ANY(col4) ;
END;
$$
LANGUAGE 'plpgsql';
CREATE FUNCTION
native=> select find_val('yo');
ERROR: structure of query does not match function result type
DETAIL: Returned type json does not match expected type integer in column 1.
CONTEXT: PL/pgSQL function find_val(text) line 3 at RETURN QUERY
native=> drop function find_val(text);
DROP FUNCTION
native=> CREATE FUNCTION find_val(val text)
native-> RETURNS json AS
native-> $$
native$> BEGIN
native$> SELECT row_to_json(sample) FROM sample where $1 = ANY(col4);
native$> END;
native$> $$
native-> LANGUAGE 'plpgsql';
CREATE FUNCTION
native=> select find_val('yo');
ERROR: query has no destination for result data
HINT: If you want to discard the results of a SELECT, use PERFORM instead.
CONTEXT: PL/pgSQL function find_val(text) line 3 at SQL statement
native=>
This is nothing to do with json vs other return types. You can't use plain SELECT
in a PL/PgSQL function, it has to be SELECT INTO
, RETURN QUERY SELECT
, or PERFORM
. Per the HINT
error.
In your case all you need is a plain SQL function.
CREATE FUNCTION find_val(val text)
RETURNS json AS
$$
SELECT row_to_json(sample) FROM sample where $1 = ANY(col4);
$$ LANGUAGE sql;
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