1

我正在尝试在 postgres 函数中使用 Jsonb。我无法正确发送参数。

mac=# CREATE TABLE json_test (id serial, json jsonb);
CREATE TABLE
mac=# INSERT INTO json_test (json) VALUES ('{"key": "value"}');
INSERT 0 1
mac=# SELECT * FROM json_test;
 id |       json       
----+------------------
  1 | {"key": "value"}
(1 row)

mac=# SELECT * FROM json_test WHERE json->'key' @> '"value"';
 id |       json       
----+------------------
  1 | {"key": "value"}
(1 row)

mac=# CREATE OR REPLACE FUNCTION testando() RETURNS setof int AS $$
mac$#   SELECT id FROM json_test WHERE json->'key' @> '"value"';
mac$# $$ LANGUAGE SQL;
CREATE FUNCTION
mac=# SELECT * FROM testando();
 testando 
----------
    1
(1 row)

mac=# CREATE OR REPLACE FUNCTION testando(value_param varchar) RETURNS setof int AS $$
mac$#   SELECT id FROM json_test WHERE json->'key' @> '"$1"';
mac$# $$ LANGUAGE SQL;
CREATE FUNCTION
mac=# SELECT * FROM testando('value');
 testando 
----------
(0 rows)

此查询应返回值:

SELECT * FROM testando('value');

有谁知道在这种情况下如何正确发送参数?

4

1 回答 1

1

编写时json->'key' @> '"$1"';,您将'"$1"'其用作带有value $1的文字常量。不要将参数括在引号中,以便它实际引用参数:

CREATE OR REPLACE FUNCTION testando(value_param varchar) RETURNS setof int AS $$
SELECT id FROM json_test WHERE json->'key' @> $1::jsonb;
$$ LANGUAGE SQL;
SELECT * FROM testando('"value"');
于 2015-04-07T09:15:43.770 回答