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:所有DISTINCT聚合函数都需要具有与count相同的参数集(DISTINCT color_key); 偏差函数:count(DISTINCT id)

我不知道为什么会出现这个错误.我所做的是为每个小组(按分组upload_key),我试图计算多少distinct id以及多少distinct color_key.
有谁有想法吗

Answer 1

错误消息表明DISTINCT只允许在一个列[组合]上,但您尝试两个,color_key&id.解决方法是两个选择,然后是一个连接:

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