如何解析以下 JSON 以检索 snowflake sql 中的项目名称？

Question

 [
  {
    "itemId": "HWKDVCXKU5",
    "name": "A",
    "quantity": 1.000000000000000e+00,
    "storeId": 1.150192000000000e+06,
    "type": "structure"
  },
  {
    "itemId": "FPK81M587X",
    "name": "b",
    "quantity": 1.000000000000000e+00,
    "storeId": 1.150192000000000e+06,
    "type": "structure"
  },
  {
    "itemId": "L04WBQON3C",
    "name": "C",
    "quantity": 2.000000000000000e+00,
    "storeId": 1.150192000000000e+06,
    "type": "structure"
  },
  {
    "itemId": "ULZFPY2UJN",
    "name": "D",
    "quantity": 2.000000000000000e+00,
    "storeId": 1.150192000000000e+06,
    "type": "structure"
  },

我想要 2 列中的所有名称和数量，我不知道如何编写相同的查询。

Answer 1

所以使用 sub-select 来“制作数据”，然后使用 PARSE_JSON 和 FLATTEN 我们得到：

SELECT 
  f.value:name::text as name,
  f.value:quantity::float as quanitiy
  FROM (
      SELECT PARSE_JSON('[
  {
    "itemId": "HWKDVCXKU5",
    "name": "A",
    "quantity": 1.000000000000000e+00,
    "storeId": 1.150192000000000e+06,
    "type": "structure"
  },
  {
    "itemId": "FPK81M587X",
    "name": "b",
    "quantity": 1.000000000000000e+00,
    "storeId": 1.150192000000000e+06,
    "type": "structure"
  },
  {
    "itemId": "L04WBQON3C",
    "name": "C",
    "quantity": 2.000000000000000e+00,
    "storeId": 1.150192000000000e+06,
    "type": "structure"
  },
  {
    "itemId": "ULZFPY2UJN",
    "name": "D",
    "quantity": 2.000000000000000e+00,
    "storeId": 1.150192000000000e+06,
    "type": "structure"
  }]') as json
       ) j
  ,TABLE(FLATTEN(input=>json)) f;

给出：

NAME	QUANITIY
A	1
b	1
C	2
D	2

查看完全相同代码但 JSON 移至 CTE 的另一种方式（因此它看起来像正常的 table）


WITH data_cte AS (
    SELECT PARSE_JSON('[
  {
    "itemId": "HWKDVCXKU5",
    "name": "A",
    "quantity": 1.000000000000000e+00,
    "storeId": 1.150192000000000e+06,
    "type": "structure"
  },
  {
    "itemId": "FPK81M587X",
    "name": "b",
    "quantity": 1.000000000000000e+00,
    "storeId": 1.150192000000000e+06,
    "type": "structure"
  },
  {
    "itemId": "L04WBQON3C",
    "name": "C",
    "quantity": 2.000000000000000e+00,
    "storeId": 1.150192000000000e+06,
    "type": "structure"
  },
  {
    "itemId": "ULZFPY2UJN",
    "name": "D",
    "quantity": 2.000000000000000e+00,
    "storeId": 1.150192000000000e+06,
    "type": "structure"
  }]') as json
)
SELECT 
  f.value:name::text as name,
  f.value:quantity::float as quanitiy
  FROM data_cte j
  ,TABLE(FLATTEN(input=>json)) f;

Answer 2

这是我解决你的问题的方法：

create or replace stage my_json_stage;
PUT file:///Users/athakur/Downloads/Chrome Downloads/test.json @my_json_stage auto_compress=false;

create or replace table test_json
( my_value variant);

copy into test_json from @my_json_stage/test.json
file_format = (type = JSON strip_outer_array = true)
ON_Error = Continue;

select * from test_json;
select :name::string, :quantity::string from test_json;

如何解析以下 JSON 以检索 snowflake sql 中的项目名称？

How to parse the following JSON to retrieve the name of the item in snowflake sql?

sql

json

snowflake-cloud-data-platform