如何在 Snowflake 中读取 JSON 键值作为数据列？

Question

我有以下示例 JSON：

{
   "Id1": {
      "name": "Item1.jpg",
      "Status": "Approved"
   },
   "Id2": {
      "name": "Item2.jpg",
      "Status": "Approved"
   }
}

我正在尝试获得以下输出：

_key    name        Status
Id1     Item1.jpg   Approved
Id2     Item2.jpg   Approved

有什么方法可以使用 SQL 在 Snowflake 中实现此目的吗？

Answer 1

您应该在保存 JSON 数据的任何列中使用 Snowflake 的 VARIANT 数据类型。让我们一步步分解：

create temporary table FOO(v variant); -- Temp table to hold the JSON. Often you'll see a variant column simply called "V"

-- Insert into the variant column. Parse the JSON because variants don't hold string types. They hold semi-structured types.
insert into FOO select parse_json('{"Id1": {"name": "Item1.jpg", "Status": "Approved"}, "Id2": {"name": "Item2.jpg", "Status": "Approved"}}');

-- See how it looks in its raw state
select * from FOO;

-- Flatten the top-level JSON. The flatten function breaks down the JSON into several usable columns
select * from foo, lateral flatten(input => (foo.v)) ;

-- Now traverse the JSON using the column name and : to get to the property you want. Cast to string using ::string.
-- If you must have exact case on your column names, you need to double quote them.
select  KEY                     as "_key",
        VALUE:name::string      as "name",
        VALUE:Status::string    as "Status"
from FOO, lateral flatten(input => (FOO.V)) ;