sql - Impala:所有 DISTINCT 聚合函数都需要具有相同的参数集

Question

我的 Impala 查询中出现以下错误:

select 
   upload_key, 
   max(my_timestamp) as upload_time, 
   max(color_key) as max_color_fk, 
   count(distinct color_key) as color_count, 
   count(distinct id) as toy_count 
from upload_table 
group by upload_key

并得到错误:

AnalysisException: all DISTINCT aggregate functions need to have the same set of parameters as count(DISTINCT color_key); deviating function: count(DISTINCT id)

我不确定为什么会出现此错误。我所做的是对每个组(按 upload_key 分组)，我试图计算有多少 distinct id以及多少 distinct color_key .
有谁有想法吗

Answer 1

错误消息表明 DISTINCT只允许一列[组合]，但你尝试两列，color_key & id .解决方法是两个 Selects 然后一个 join:

select
   t1.upload_key,
   t1.upload_time,
   t1.max_color_fk,
   t1.color_count,
   t2.toy_count
from
 (
   select 
      upload_key, 
      max(my_timestamp) as upload_time, 
      max(color_key) as max_color_fk, 
      count(distinct color_key) as color_count
   from upload_table 
   group by upload_key
 ) as t1
join
 (
   select 
      upload_key
      count(distinct id) as toy_count 
   from upload_table 
   group by upload_key
 ) as t2
on t1. upload_key = t2.upload_key