我了解到可以使用row_to_json来返回json格式的输出。
例如,如果我的查询是:
例如,如果我的查询是:
select * from sample;
我可以将其重写为以下形式,以返回json输出:
select row_to_json(sample) from sample;
但我想要实现的一件事情是在函数中实现相同的功能。
为了给你举个例子,这里有一个返回表格的函数:
CREATE FUNCTION find_val(val text)
RETURNS SETOF sample AS
$$
BEGIN
RETURN QUERY
SELECT * FROM sample where $1 = ANY(col4);
END;
$$
LANGUAGE 'plpgsql';
现在我想从我的函数返回 JSON 输出,而不是行。我该怎么做呢?
到目前为止,我尝试了以下方法:
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=>
创建或替换函数 find_val(val text) 返回 json AS $$ DECLARE t_row sample%ROWTYPE; BEGIN SELECT * INTO t_row FROM sample where $1 = ANY(col4); RETURN row_to_json(t_row); END; $$ 语言 'plpgsql';
- Mandeep Singh