Question

我已阅读related question，但与用户kenthewala不同，我希望将JSON对象数组添加到数据库中。

我的JSON文件如下所示：

{
"tablename_a":[{"a_id":1,"b_id":2,"c_id":3},
 {"a_id":2,"b_id":51,"c_id":3}],
"tablename_b":[{"b_id":2,"name":"John Doe", "z_id":123},
 {"b_id":51,"name":"Mary Ann", "z_id":412}],
"tablename_c":[{"c_id":3, "OS type":"Windows 7"}],
"tablename_z":[{"z_id":123, "Whatever":"Something"},
{"z_id":123, "Whatever":"Something else"}]
}

db。中已存在具有相应名称的表。

在伪代码中我想象了类似

for each key in JSON_FILE as tbl_name
(
  insert into tbl_name select * from json_populate_recordset
  (
    null::tbl_name, 'content of tbl_name'
  )
)

但我不确定，如何实现这一点。

我正在使用PostgreSQL 9.3.5（如果有帮助的话，还有PHP 5.3.3）。

表结构类似于JSON文件（因为我最初从数据库导出JSON）：

create table tablename_a (a_id integer, b_id integer, c_id integer);
create table tablename_b (b_id integer, name text, z_id integer);

等等。

Answer 1

3个步骤：

访问JSON对象的字段->。
使用json_populate_recordset()从JSON记录数组创建派生表。
分解INSERT命令的行类型。

要重用所有表的输入值，请将其包装在data-modifying CTEs：

WITH input AS (
   SELECT '{
      "tablename_a":[{"a_id":1,"b_id":2,"c_id":3},
       {"a_id":2,"b_id":51,"c_id":3}],
      "tablename_b":[{"b_id":2,"name":"John Doe", "z_id":123},
       {"b_id":51,"name":"Mary Ann", "z_id":412}],
      "tablename_c":[{"c_id":3, "OS type":"Windows 7"}],
      "tablename_z":[{"z_id":123, "Whatever":"Something"},
      {"z_id":123, "Whatever":"Something else"}]
      }'::json AS j
   )
,  a AS (
   INSERT INTO tablename_a
   SELECT t.*
   FROM   input i
        , json_populate_recordset(NULL::tablename_a, i.j->'tablename_a') t
   )
,  b AS (
   INSERT INTO tablename_b
   SELECT t.*
   FROM   input i
        , json_populate_recordset(NULL::tablename_b, i.j->'tablename_b') t
   )
   -- ... more ...
INSERT INTO tablename_z
SELECT t.*
FROM   input i
     , json_populate_recordset(NULL::tablename_z, i.j->'tablename_z') t
;

<强> SQL Fiddle.

使用隐式JOIN LATERAL。相关：

Query combinations with nested array of records in JSON datatype

使用json_populate_recordset从单个JSON对象填充多个表

1 个答案: