GreenPlum SQL 查询转换成 Hive

Posted 2023-03-23

【中文标题】GreenPlum SQL 查询转换成 Hive

【英文标题】：GreenPlum SQL query to convert into Hive

【发布时间】：2021-11-08 09:43:34

【问题描述】：

下面的查询我们想将greenplum SQL转换为hive sql，请帮助我们。

GP子查询：

select a.region , amount_fr , count_fr , amount_sr , count_sr from (
select region , sum(cast(amount as integer)) as amount_fr , count(transid) as count_fr  from test.test_fr_imi
where cast(trans_date as date) between (select cast(add_months(trunc(date_sub(cast(current_date as date),1),'MM'),-1) as date) ) and
(select current_date::date  - 1 - substring((select (date_trunc('month',current_date::date ))::date - 1)::character varying,9,10)::integer)
  group by 1 ) a
JOIN
(select region , sum(amount::integer) as amount_sr , count(transid) as count_sr from test.test_sr_imi
where trans_date::date between (select (date_trunc('month',current_date::date - 1) - interval '1 month')::date) and
(select current_date::date  - 1 - substring((select (date_trunc('month',current_date::date ))::date - 1)::character varying,9,10)::integer) group by 1 ) b on
a.region = b.region;

我需要转换上面的查询，子查询不能转换成hive，请找到上面的特定代码。

(select current_date::date - 1 - substring((select (date_trunc('month',current_date::date))::date - 1)::character 变化的,9,10)::整数)

select to_char(round((select sum(revenue)/1000000.00 from test.sampletable where trxn_date = current_date-1)/1, 2) ,'999,999')

select to_char(round((select sum(case when to_char(trxn_date, 'yyyymm') = to_char((current_date - 1) - '1 month'::interval, 'yyyymm') and extract(day from trxn_date) < extract(day from current_date) then revenue end)/1000000.00 from test.sample2table)/1, 2) ,'999,999')

select to_char(round ((select(( ( sum(case when to_char(trxn_date, 'yyyymm') = to_char((current_date - 1), 'yyyymm') then revenue end))  -   (sum(case when to_char(trxn_date, 'yyyymm') = to_char((current_date - 1) - '1 month'::interval, 'yyyymm') and extract(day from trxn_date) < extract(day from current_date)  then revenue end) )) / (sum(case when to_char(trxn_date, 'yyyymm') = to_char((current_date - 1) - '1 month'::interval, 'yyyymm') and extract(day from trxn_date) < extract(day from current_date) then revenue end) ))*100.00 from test.sample3table)/1, 2) ,'9990.99%')

【参考方案1】：

这里有一些关于某些功能的帮助-

round() - 在 hive 中可用。 to_char(INT,'999,999') - 您可以使用 format_number(12345,0) - 这将产生 12,345 CURRENT_DATE- 在 hive 中可用。 CURRENT_DATE-1 - 这可以使用 current_date - interval '1' day 完成。 hive 中的语法略有不同。您需要将您的 sql 放在子查询中并在其上进行计算。

所以，这里相当于第一个sql-

SELECT format_number(round(sum_rev/1, 2), 0)
FROM 
    (SELECT sum(revenue)/1000000.00 as sum_rev
     FROM test.sampletable
     WHERE trxn_date = CURRENT_DATE- interval '1' day) rs 

这里是第二个查询的等效 sql

SELECT format_number(round(sum_rev/1, 2), 0) 
FROM (SELECT sum(CASE
                                       WHEN from_timestamp(trxn_date, 'yyyyMM') = from_timestamp((CURRENT_DATE - interval '1' day) - interval '1' month, 'yyyyMM')
                                            AND extract(DAY
                                                        FROM trxn_date) < extract(DAY
                                                                                  FROM CURRENT_DATE) THEN revenue
                                   END)/1000000.00 as sum_rev
                        FROM test.sample2table) rs  

【讨论】：

(select current_date::date - 1 - substring((select (date_trunc('month',current_date::date))::date - 1)::character varying,9,10): :integer)

我们如何用子查询进行转换

select cast(date_sub(cast(current_date as date),1) - substring((select cast(date_sub(cast(trunc(cast(current_date as date),'MM')as date), 1) 作为字符串),9,10)) 作为 bigint)

我不明白这一点。查询完全改变了。同样，我可以在同一个子查询上看到另一个问题。你能专注于原始查询及其答案吗？

@karthikelavan 在有人帮助你并且你说它有效之后，请避免编辑最初的问题。问另一个问题并考虑接受/支持这个问题。当您不断编辑问题时，这是在浪费我们的时间。如果它有效，请不要忘记接受/投票，这就是在这个网站上说“谢谢它有效”的方式。