运算符不存在 - Postgres & JSON Select 查询

Question

我正在尝试从包含 JSONB 数据的 table 中检索和连接数据，其中 JSON 行的格式为：

{
    "id": "d57929b35216",
    "base" : {"legalName" : "SAPI S.P.A."}, 
    "name": "SAPI S.P.A.", 
}

Table ita_public

CREATE TABLE public.ita_public
(
    id integer NOT NULL DEFAULT nextval('ita_data_id_seq'::regclass),
    info jsonb NOT NULL,
    CONSTRAINT ita_data_pkey PRIMARY KEY (id)
)

Table ita_sn_private

CREATE TABLE public.ita_sn_private
(
    id integer NOT NULL DEFAULT nextval('ita_sn_private_id_seq'::regclass),
    supplier_name character varying COLLATE pg_catalog."default",
    supplier_streetadd character varying COLLATE pg_catalog."default",
    CONSTRAINT ita_sn_private_pkey PRIMARY KEY (id)
)

SELECT 查询打印三列，加入供应商名称和嵌套名称并搜索名称：

SELECT
priv.supplier_name,
priv.supplier_streetadd,
pub.info::json->'base'->'legalName'

FROM ita_sn_private as priv 
JOIN ita_public as pub ON (priv.supplier_name = pub.info::json->'name')

WHERE pub.info::json->>'name' = 'SAPI S.P.A.'

我收到错误：

ERROR: operator does not exist:

character varying = json LINE 7: JOIN ita_public as pub ON (priv.supplier_name = pub.info::js...

HINT: No operator matches the given name and argument type(s). You might need to add explicit type casts.

('='下面的小帽子)

我试图简化查询以确保我的 json 路径是正确的：

SELECT
info::json->'base'->'legalName'
FROM
ita_public
WHERE info::json->>'name' = 'SAPI S.P.A.'

效果很好。

任何人都可以协助 JOIN 语句吗？我不确定如何将两者等同起来。

Answer 1

加入条件是原因：

on priv.supplier_name = pub.info::json->'name'

-> returns 一个 JSON 对象，不是字符串。你需要在那里使用 ->>

on priv.supplier_name = pub.info::json->>'name'

与您在 WHERE 子句中所做的相同

运算符不存在 - Postgres & JSON Select 查询

Operator Does Not Exist - Postgres & JSON Select Query

postgresql

json

jsonb