將 JSON 值列表分解為 SNOWFLAKE 數據庫表中的行

Question

我按照下面的屏幕截圖有一個表格，它基本上是一個 JSON 並被解析以獲得下面的輸出，現在我想要列 City 和 orders 中的值列表被拆分成行。

有人可以幫我嗎

所需的輸出如下

Answer 1

這是一種方法。 首先去掉 [" 和 ]" 因為 city 列中的雙引號不包含單個數組元素而是包含所有元素，然后將字符串標記化並使用 strtok_to_array 將其作為真實數組返回，然后將數組元素展平以分隔行並將行（城市）橫向連接回記錄的其余部分。

with data as
(select 'A' as name, 'M' as gender, '["completed"]' as orders, '["Cochi,Hyderabad"]' as city
union all
 select 'B' as name, 'M' as gender, '["completed"]' as orders, '["Cochi,Hyderabad,Delhi"]' as city
union all
 select 'C' as name, 'F' as gender, '["cancelled"]' as orders, '["Mumbai,Pune"]' as city
union all
 select 'D' as name, 'M' as gender, '["pending"]' as orders, '["cochi"]' as city
)
, data2 as 
( select d.name
 , d.gender
 , replace(replace(d.orders,'["',''),'"]','') as orders
 , strtok_to_array(replace(replace(city,'["',''),'"]',''),',')  as city
 from data d
)
 select d2.name
 , d2.gender
 , d2.orders
 , replace(c.value,'"','') as city
 from data2 d2
  , lateral flatten(input => d2.city) c;

Answer 2

如果city字段中的數組實際上是一個包含單個逗號分隔字符串的數組，如["Cochi, Hyderabad"] ，則需要使用STRTOK_TO_ARRAY進行第三級展平，如下所示：

create or replace table t1 (json variant) as
select parse_json('{"Name": "A", "Gender": "M", "orders": ["completed"], "city": ["Cochi, Hyderabad"]}')
union all
select parse_json('{"Name": "B", "Gender": "M", "orders": ["completed"], "city": ["Cochi, Hyderabad, Delhi"]}')
union all
select parse_json('{"Name": "C", "Gender": "F", "orders": ["cancelled"], "city": ["Mumbai, Pune"]}')
union all
select parse_json('{"Name": "D", "Gender": "M", "orders": ["pending"], "city": ["Cochi"]}')
;

select
    json:Name::varchar name,
    json:Gender::varchar gender,
    orders.value::varchar orders,
    city.value::varchar city
from t1,
lateral flatten(json:orders) orders,
lateral flatten(json:city) city_raw,
lateral flatten(strtok_to_array(city_raw.value, ', ')) city
;
/*
NAME    GENDER  ORDERS  CITY
A   M   completed   Cochi
A   M   completed   Hyderabad
B   M   completed   Cochi
B   M   completed   Hyderabad
B   M   completed   Delhi
C   F   cancelled   Mumbai
C   F   cancelled   Pune
D   M   pending Cochi
*/

LATERAL FLATTEN是一種將對象 (JSON) 或數組中的值展平（擴展）為行並與父表中的原始行組合的習慣用法。

所以，上面的查詢是：

• 展平orders字段中的數組以將數組元素展開為輸出中ORDERS列的行
• 將city字段中的數組展平，將數組元素展開為輸出中CITY_RAW中間列的行
• 將CITY_RAW列中提取的逗號分隔字符串拆分為城市名稱，並將字符串的中間數組展平為CITY列中的行

這種方式使您無需任何復雜的字符串操作即可獲得所需的輸出。

---

但是，我懷疑存儲在 JSON 中的數組實際上就像["Cochi", "Hyderabad"] ，每個城市名稱用雙引號引用的城市名稱數組。

在這種情況下，查詢將很簡單； 您只需要通過對每個字段使用LATERAL FLATTEN來展平orders字段和city字段中的數組。

create or replace table t2 (json variant) as
select parse_json('{"Name": "A", "Gender": "M", "orders": ["completed"], "city": ["Cochi", "Hyderabad"]}')
union all
select parse_json('{"Name": "B", "Gender": "M", "orders": ["completed"], "city": ["Cochi", "Hyderabad", "Delhi"]}')
union all
select parse_json('{"Name": "C", "Gender": "F", "orders": ["cancelled"], "city": ["Mumbai", "Pune"]}')
union all
select parse_json('{"Name": "D", "Gender": "M", "orders": ["pending"], "city": ["Cochi"]}')
;


select
    json:Name::varchar name,
    json:Gender::varchar gender,
    orders.value::varchar orders,
    city.value::varchar city
from t2,
lateral flatten(json:orders) orders,
lateral flatten(json:city) city
;
/*
NAME    GENDER  ORDERS  CITY
A   M   completed   Cochi
A   M   completed   Hyderabad
B   M   completed   Cochi
B   M   completed   Hyderabad
B   M   completed   Delhi
C   F   cancelled   Mumbai
C   F   cancelled   Pune
D   M   pending Cochi
*/