hive ParseException行无法识别joinSource中“(”“select”“from”附近的输入

dxpyg8gm 于 2022-11-05 发布在 Hive

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我试图在配置单元中执行查询，但出现错误。我一遍又一遍地检查，但看不到任何问题。

select 
a.phone_no,
a.app_name
from 
(select * from (select app_name,phone_no from lc_app_flag) ) a 
inner join 
(select * from (select phone_no,city_id_day,city_id_night,lat_day,lng_day,lat_night,lat_night from TW_FEATS_FIN_LCEXT01 where month_id='202205'  ) b where city_id_day='440100' or city_id_night='440100') c
on a.phone_no=c.phone_no

Hive

来源：https://stackoverflow.com/questions/72882971/parseexception-line-cannot-recognize-input-near-select-from-in-joinsourc

1条答案

按热度按时间

lnvxswe21#

您缺少第一个最内层子查询别名请更正下面SQL-

select 
a.phone_no,
a.app_name
from 
(select * from (select app_name,phone_no from lc_app_flag) subq)  a --inner subquery as subq
inner join 
(select * from (select phone_no,city_id_day,city_id_night,lat_day,lng_day,lat_night,lat_night from TW_FEATS_FIN_LCEXT01 where month_id='202205'  ) b where city_id_day='440100' or city_id_night='440100') c
on a.phone_no=c.phone_no

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hive ParseException行无法识别joinSource中“(”“select”“from”附近的输入

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