我试图在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');
在这种情况下,有谁知道如何正确发送参数?
答案 0 :(得分:1)
撰写json->'key' @> '"$1"';
时,您使用'"$1"'
作为值 $1
的文字常量。不要将参数包装在引号中,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"');