使用hql将数据从一个表转换到另一个表

3zwtqj6y  于 2021-06-01  发布在  Hadoop
关注(0)|答案(1)|浏览(374)

我现在有一个类似这样的表:
表1

ID|NAME|INFO_CODE_1|INFO_CODE_2|INFO_CODE_3|INFO_CODE_4|INFO_TEXT|DESCRIPTION
1 |Test|123        |254        |556        |867        |Test Text|Test Description

我试图将表1中的数据插入表2中,这样第一个表中的每条记录都会变成第二个表中的四条独立记录,如下所示:
表2

ID|NAME|INFO_CODE|INFO_TEXT|DESCRIPTION
1 |Test|123      |Test Text|Test Description
1 |Test|254      |Test Text|Test Description
1 |Test|556      |Test Text|Test Description
1 |Test|867      |Test Text|Test Description

表\u 1中有成千上万条记录,还有其他列的数据我们不关心,而info\u code \u 1到info\u code \u 4中的值都因记录而异(甚至可能为空)。我在考虑写这样的东西:

INSERT OVERWRITE TABLE example.TABLE_2
SELECT
A.ID AS ID,
A.NAME AS NAME,
A.INFO_CODE_1 AS INFO_CODE,
A.INFO_TEXT AS INFO_TEXT,
A.DESCRIPTION AS DESCRIPTION
FROM
example.TABLE_1 A
UNION ALL
SELECT
A.ID AS ID,
A.NAME AS NAME,
A.INFO_CODE_2 AS INFO_CODE,
A.INFO_TEXT AS INFO_TEXT,
A.DESCRIPTION AS DESCRIPTION
FROM
example.TABLE_1 A
UNION ALL
SELECT
A.ID AS ID,
A.NAME AS NAME,
A.INFO_CODE_3 AS INFO_CODE,
A.INFO_TEXT AS INFO_TEXT,
A.DESCRIPTION AS DESCRIPTION
FROM
example.TABLE_1 A
UNION ALL
SELECT
A.ID AS ID,
A.NAME AS NAME,
A.INFO_CODE_4 AS INFO_CODE,
A.INFO_TEXT AS INFO_TEXT,
A.DESCRIPTION AS DESCRIPTION
FROM
example.TABLE_1 A;

这样行吗?
如果是的话,有没有更好的方式写声明?
有没有办法确保具有相同id的所有记录都被依次插入?
提前感谢你们能提供的任何帮助。

vsdwdz23

vsdwdz231#

另一个没有太多工会的办法是这样。它将需要1个mapreduce作业,而不是几个节省资源的作业。您可以将代码合并、拆分以生成数组并分解数组。

INSERT OVERWRITE TABLE example.TABLE_2
SELECT
a.id,
a.name,
CODE
a.INFO_TEXT,
A.DESCRIPTION from 
(
SELECT
A.ID AS ID,
A.NAME AS NAME,
split(concat(A.INFO_CODE_1, ",",A.INFO_CODE_2,",",A.INFO_CODE_3,",",A.INFO_CODE_4),",") AS INFO_CODE_ARRAY,
A.INFO_TEXT AS INFO_TEXT,
A.DESCRIPTION AS DESCRIPTION
from TABLE_1
) t LATERAL VIEW explode(INFO_CODE_ARRAY) codea AS code;

相关问题