如何从多个配置单元表创建新的配置单元表？我使用JSON字符串创建原始表，但不确定如何创建新表

如何从多个配置单元表创建新的配置单元表？我使用JSON字符串创建原始表，但不确定如何创建新表

create table Bank_F001_table as
select
    get_json_object(Bank_F001.json, '$.text') as text,
    get_json_object(Bank_F001.json, '$.coordinates') as coordinates,
    get_json_object(Bank_F001.json, '$.user.location') as location,
    get_json_object(Bank_F001.json, '$.lang') as lang,
    get_json_object(Bank_F001.json, '$.user_mentions') as user_mentions,
    get_json_object(Bank_F001.json, '$.user.screen_name') as screen_name,
    get_json_object(Bank_F001.json, '$.user.name') as name,
    get_json_object(Bank_F001.json, '$.listed_count') as listed_count,
    get_json_object(Bank_F001.json, '$.in_reply_to_user_id_str') as in_reply_to_user_id_str
from Bank_F001;

只是想添加，我使用UNION ALL来选择成功运行的属性。但我无法创建用于其他活动的新表。奇怪的是，我也尝试了UNION（而不是ALL），Hive告诉我需要使用ALL。 这是我的选择代码

select *
from BMO_F016_table
union all
select *
from BMO_F017_table
union all
select *
from BMO_F018_table
union all
select *
from BMO_F054_table ) as large_table
group by screen_name
order by cnt desc;

---

出于某种原因，它不起作用，但我发现了。。。如果你感兴趣，给你

CREATE TABLE BMO_Table as
SELECT *
FROM BMO_F016_table
UNION ALL
SELECT *
FROM BMO_F017_table
UNION ALL
SELECT *
FROM BMO_F018_table
UNION ALL
SELECT *
FROM BMO_F054_table
UNION ALL
SELECT *
FROM BMO_F093_table
;

---

出于某种原因，它不起作用，但我发现了。。。如果你感兴趣，给你

CREATE TABLE BMO_Table as
SELECT *
FROM BMO_F016_table
UNION ALL
SELECT *
FROM BMO_F017_table
UNION ALL
SELECT *
FROM BMO_F018_table
UNION ALL
SELECT *
FROM BMO_F054_table
UNION ALL
SELECT *
FROM BMO_F093_table
;