Question

我有一张桌子：

Sales ID
100   a
100   a
103.5 c
105   d
105   d

... 我只想获取值 308.5 （即100 + 103.5 + 105，唯一值）。

我已经尝试过了：

select sum(case when rownum=1 then sales end) from 
(select sales,id, row_number() over (partition  by id) as rownum from table) subquery

但是每次刷新时，我都会得到不同的值。

替代方法是什么？

Answer 1

但是每次刷新时，我都会得到不同的值。

这表明基础数据正在更改-这在Vertica上不太可能。或更可能您对一个ID具有多个值。您可以测试后者：

select id, count(distinct sales)
from t
where min(sales) <> max(sales);

Answer 2

您可以将子查询用于不同的销售

select sum(quantity), count(*), . . .
from orders o
where not exists (select 1 from orders o2 where o2.customercd = o.customercd and o2.type = 1);

Answer 3

您必须先运行SELECT DISTINCT，然后再运行SUM（）-ming来进行重复数据删除。像这样：

WITH
-- this is your input - don't use in real query
input(Sales,ID) AS (
          SELECT 100,'a'
UNION ALL SELECT 100,'a'
UNION ALL SELECT 103.5,'c'
UNION ALL SELECT 105,'d'
UNION ALL SELECT 105,'d'
)
, -- < - replace this comma with "WITH", and start the real query here ...
uq (sales) AS (
  SELECT DISTINCT
    sales
  FROM input
)
SELECT
  SUM(sales)
FROM uq;
-- out   SUM  
-- out -------
-- out  308.5
-- out (1 row)
-- out 
-- out Time: First fetch (1 row): 7.795 ms. All rows formatted: 7.826 ms


-- if ( 100 , 'a' ) is different from ( 100 , 'd' ), then it would be the below

WITH
-- this is your input - don't use in real query
input(Sales,ID) AS (
          SELECT 100,'a'
UNION ALL SELECT 100,'a'
UNION ALL SELECT 103.5,'c'
UNION ALL SELECT 105,'d'
UNION ALL SELECT 105,'d'
)
, -- < - replace this comma with "WITH", and start the real query here ...
uq (sales,id) AS (
  SELECT DISTINCT
    sales
  , id
  FROM input
)
SELECT
  SUM(sales)
FROM uq;
-- out   SUM  
-- out -------
-- out  308.5
-- out (1 row)
-- out 
-- out Time: First fetch (1 row): 11.084 ms. All rows formatted: 11.117 ms

Answer 4

您想要来自不同的销售对和ID的销售总和：

select sum(t.sales) from (
  select distinct sales, id
  from table
) t

或：

select sum(t.sales) from (
  select sales from table
  group by sales, id
) t

基于另一列的不同值的SUM值

4 个答案: