在hive-rank()中每天获取前n行

dbf7pr2w  于 2021-06-26  发布在  Hive
关注(0)|答案(2)|浏览(275)

我有一张table,每一排都有一笔交易:

sale_date  salesman  sale_item_id
 20170102   JohnSmith       309
 20170102   JohnSmith       292
 20170103   AlexHam          93

我每天都想找到前20名推销员,我想到了这个:

SELECT sale_date, salesman, sale_count, row_num
FROM (
  SELECT sale_date, salesman,
         count(*) as sale_count,
         rank() over (partition by sale_date order by sale_count desc) as row_num
  from salesforce.sales_data
) T
WHERE sale_date between  '20170101' and '20170110'
 and row_num <= 20

但我得到:

FAILED: SemanticException Failed to breakup Windowing invocations into Groups. At least 1 group must only depend on input columns. Also check for circular dependencies.
Underlying error: org.apache.hadoop.hive.ql.parse.SemanticException: Line 5:35 Expression not in GROUP BY key 'sale_date'

不过,我不确定分组会在什么时候生效。有人能帮忙吗?德克萨斯州!

jogvjijk

jogvjijk1#

试试这个

SELECT sale_date, salesman, sale_count, row_num from (
SELECT sale_date, salesman, sale_count,
 rank() over (partition by sale_date order by sale_count desc) as         row_num
from 
(
SELECT sale_date, salesman,
   count(*) over (partition by salesman) as sale_count
from  employee
) t1
) t2  where sale_date between  '20170101' and '20170110'
and row_num <= 20;
WHERE sale_date between  '20170101' and '20170110'
and row_num <= 20

编辑和测试。您的问题本质上是,您试图在计算over子句之前使用count,如果您在子查询pariting by sallers中计算count,它将解决问题。您不能在销售查询中执行分组依据,如果执行,您将无权访问销售日期。

1tuwyuhd

1tuwyuhd2#

你错过了一个 group by 在子查询中:

SELECT sale_date, salesman, sale_count, row_num
FROM (SELECT sale_date, salesman,
             count(*) as sale_count,
             rank() over (partition by sale_date order by count(*) desc) as row_num
      FROM salesforce.sales_data
      GROUP BY sale_date, salesman
     ) T
WHERE sale_date between '20170101' and '20170110' and row_num <= 20;

我认为hive将接受 order by , order by sale_count desc .
另外请注意,如果有关系,则可以获得多于或少于20行的数据。你可能想要 row_number() 如果你正好需要20排。

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