展平Postgres嵌套JSONB列

Question

我希望了解如何扁平化嵌套在JSONB列中的数据。例如，假设我们有表users和user_id(int)和同级表(JSONB)

其中的行如下：

id | JSONB
---------------------
1  | {"brother": {"first_name":"Sam", "last_name":"Smith"}, "sister": {"first_name":"Sally", "last_name":"Smith"}
2  | {"sister": {"first_name":"Jill"}}

我正在寻找一个将返回如下响应的查询：

id | sibling   | first_name | last_name
-------------------------------------
1  | "brother" | "Sam"      | "Smith"
1  | "sister"  | "Sally"    | "Smith"
2  | "sister"  | "Jill"     | null

Answer 1

我在psql中使用它来开发它。为了检查代码，我创建了一个小视图t1

CREATE VIEW t1 AS (
       SELECT 1 AS id, '{"brother": {"first_name":"Sam", "last_name":"Smith"}, "sister": {"first_name":"Sally", "last_name":"Smith"}}'::jsonb AS jsonb
 UNION SELECT 2, '{"sister": {"first_name":"Jill", "last_name":"Johnson"}}'
 UNION SELECT 3, '{"sister": {"first_name":"Jill", "x_name":"Johnson"}}'
);

第一个任务是找到可能的关键字列表：

WITH fields AS (
     SELECT DISTINCT jff.key
       FROM t1,
            jsonb_each(jsonb) AS jf,
            jsonb_each(jf.value) AS jff
)
SELECT * FROM fields;

结果是：

    key     
------------
 first_name
 last_name
 x_name

下一步是生成查询：

SELECT 'SELECT id, jf.key as sibling, ' || (
    WITH fields AS (
         SELECT DISTINCT jff.key
           FROM t1,
                jsonb_each(jsonb) AS jf,
                jsonb_each(jf.value) AS jff
    )
    SELECT string_agg('jf.value->>''' || key || ''' as "' || key || '"', ',' ORDER BY key)
      FROM fields
)
|| ' FROM t1, jsonb_each(jsonb) AS jf ORDER BY 1, 2, 3;' AS cmd;

它返回：

                                                                                  cmd                                                                                   
------------------------------------------------------------------------------------------------------------------------------------------------------------------------
 SELECT id, jf.key as sibling,jf.value->>'first_name' as "first_name",jf.value->>'last_name' as "last_name",jf.value->>'x_name' as "x_name" FROM t1, jsonb_each(jsonb) AS jf ORDER BY 1, 2, 3;
(1 row)

要将结果设置为psql变量，我使用gset

\gset

之后，您可以调用query：

:cmd

 id | sibling | first_name | last_name | x_name  
----+---------+------------+-----------+---------
  1 | brother | Sam        | Smith     | 
  1 | sister  | Sally      | Smith     | 
  2 | sister  | Jill       | Johnson   | 
  3 | sister  | Jill       |           | Johnson
(4 rows)

要从外部语言运行它，您可以创建postgres函数，然后返回SQL命令：

CREATE OR REPLACE FUNCTION build_query(IN tname text, OUT cmd text)  AS $sql$
BEGIN 
    EXECUTE $cmd$
            SELECT 'SELECT id, jf.key as sibling, ' || (
                    WITH fields AS (
                        SELECT DISTINCT jff.key
                          FROM t1,
                               jsonb_each(jsonb) AS jf,
                               jsonb_each(jf.value) AS jff
                    )
                    SELECT string_agg('jf.value->>''' || key || ''' as "' || key || '"', ',' ORDER BY key)
                      FROM fields
                )
        || ' FROM $cmd$ || quote_ident(tname) || $cmd$ , jsonb_each(jsonb) AS jf ORDER BY 1, 2, 3;'$cmd$ INTO cmd;
    RETURN;
END;
$sql$ LANGUAGE plpgsql;

SELECT * FROM build_query('t1');
                                                                                               cmd                                                                                               
-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
 SELECT id, jf.key as sibling, jf.value->>'first_name' as "first_name",jf.value->>'last_name' as "last_name",jf.value->>'x_name' as "x_name" FROM t1 , jsonb_each(jsonb) AS jf ORDER BY 1, 2, 3;
(1 row)