Question

我的要求是获取具有多个电子邮件ID且类型为1的学生的ID和姓名。

我正在使用类似的查询

select distinct b.id, b.name, b.email, b.type,a.cnt
from (
  select id, count(email) as cnt
  from (
    select distinct id, email
    from table1
  ) c
  group by id
) a
join table1 b on a.id = b.id
where b.type=1
order by b.id

请告知我这个版本是否可用或更简单。

Sample data is like:
id name email type 
123 AAA abc@xyz.com 1
123 AAA acd@xyz.com 1
123 AAA ayx@xyz.com 3
345 BBB nch@xyz.com 1
345 BBB nch@xyz.com 1
678 CCC iuy@xyz.com 1


Expected Output:
123 AAA abc@xyz.com 1 2
123 AAA acd@xyz.com 1 2
345 BBB nch@xyz.com 1 1
678 CCC iuy@xyz.com 1 1

Answer 1

您可以使用group by-> having count()来满足此要求。

select distinct b.id
    , b.name,
    , b.email
    , b.type 
from table1 b
where id in 
    (select distinct id from table1 group by email, id having count(email) > 1) 
and b.type=1 
order by b.id

Answer 2

您可以尝试使用count（）函数的解析方式：

SELECT sub.ID, sub.NAME
  FROM (SELECT ID, NAME, COUNT (*) OVER (PARTITION BY ID, EMAIL) cnt
          FROM raw.crddacia_raw) sub
 WHERE sub.cnt > 1 AND sub.TYPE = 1

Answer 3

我强烈建议您使用窗口功能。但是，Hive不支持count(distinct)作为窗口函数。有多种解决方法。一个是dense_rank()的总和：

select id, name, email, type, cnt
from (select t1.*,
             (dense_rank() over (partition by id order by email) +
              dense_rank() over (partition by id order by email desc)
             ) as cnt
      from table1 t1
     ) t
where type = 1;

我希望它具有比您的版本更好的性能。但是，值得测试不同版本以查看哪个具有更好的性能（并随时回来告诉其他人哪个更好）。

Answer 4

另一种使用collect_set并采用返回数组的大小来计算不同电子邮件的方法。

演示：

--your data example
with table1 as ( --use your table instead of this 
  select stack(6,
    123, 'AAA', 'abc@xyz.com', 1,
    123, 'AAA', 'acd@xyz.com', 1,
    123, 'AAA', 'ayx@xyz.com', 3,
    345, 'BBB', 'nch@xyz.com', 1,
    345, 'BBB', 'nch@xyz.com', 1,
    678, 'CCC', 'iuy@xyz.com', 1
  ) as (id, name, email, type )
) 

--query   
select distinct id, name, email, type, 
       size(collect_set(email) over(partition by id)) cnt
  from table1
 where type=1

结果：

id  name    email   type    cnt 
123 AAA  abc@xyz.com    1   2   
123 AAA  acd@xyz.com    1   2   
345 BBB  nch@xyz.com    1   1   
678 CCC  iuy@xyz.com    1   1

这里我们仍然需要DISTINCT，因为分析功能不会像情况345 BBB nch@xyz.com那样删除重复项。

Answer 5

这与您的查询非常相似，但在这里我要在初始步骤（内部查询）中过滤数据，以使联接不会发生在较少的数据上

 select distinct b.id,b.name,b.email,b.type,intr_table.cnt from table1 orig_table join 
    (
        select a.id,a.type,count(a.email) as cnt from table1 as a where a.type=1 group by a
    ) intr_table on inter_table.id=orig_table.id,inter_table.type=orig_table.type

配置单元查询优化

5 个答案: