在teradata sql 中划分两条记录并打印一条记录

Question

我需要从 sql 查询中的 table 中划分两条记录我在 table

中有下一条记录

表 1

Id    field1  field3
1     20      20161001  
1     35      20161101
2     20      20161001
2     30      20161101
5     19      20161001 
5     33      20161101

（请注意示例中的日期是 2016-OCT-01 和 2016-NOV-01）

我需要一个 sql 下一个结果的查询：

Id   fieldnew
1    0.75
2    0.5
5    0.7368

计算"fieldnew"的规则是： ID 1 示例

(35 ÷ 20) -1 = 0.75

字段 3 中日期为 20161101 的字段 1 ÷ 字段 3 中日期为 20161001 的字段 1 - id 为 1 的记录的 1

你能帮忙查询一下以获得所需的结果吗

Answer 1

类似这样。

with
--  start of test data (not part of the solution)
     table1 ( id, field1, field3 ) as (
       select 1, 20, 20161001 from dual union all 
       select 1, 35, 20161101 from dual union all 
       select 2, 20, 20161001 from dual union all 
       select 2, 30, 20161101 from dual union all 
       select 5, 19, 20161001 from dual union all 
       select 5, 33, 20161101 from dual
     ),
--  end of test data (not part of the solution)
     prep ( id, field1, lag_field1, field3 ) as (
       select id, field1, lag(field1) over (partition by id order by field3), field3
       from   table1
       where  field3 in (20161001, 20161101)
     )
select id, round(field1 / lag_field1, 4) - 1 as fieldnew
from   prep
where  field3 = 20161101
;

ID   FIELDNEW
--   --------
 1       0.75
 2        0.5
 5     0.7368

Answer 2

我认为以下解决了问题：

select id,
       (max(case when field3 = '20161101' then field1 end) /
        max(case when field3 = '20161001' then field1 end)
       ) - 1 as fieldnew
from table1 t1
group by id;

这是使用条件聚合来获取值。

Answer 3

甲骨文

select      id
           ,round
            (
                            min (field1) keep (dense_rank last  order by field3)
                /   nullif (min (field1) keep (dense_rank first order by field3),0)
                -   1   
               ,4
            )      as fieldnew

from        mytable

where       field3 in 
            (
                add_months (trunc (current_date,'mm'),-2)  
               ,add_months (trunc (current_date,'mm'),-1)
            ) 

group by    id
;

或

select      id
           ,round
            (
                            min (case when field3 = add_months (trunc (current_date,'mm'),-1) then field1 end)
                /   nullif (min (case when field3 = add_months (trunc (current_date,'mm'),-2) then field1 end),0)
                -   1   
               ,4
            )      as fieldnew

from        mytable

where       field3 in 
            (
                add_months (trunc (current_date,'mm'),-2)  
               ,add_months (trunc (current_date,'mm'),-1)
            ) 

group by    id
;

+----+----------+
| ID | FIELDNEW |
+----+----------+
| 1  | 0.75     |
+----+----------+
| 2  | 0.5      |
+----+----------+
| 5  | 0.7368   |
+----+----------+

天睿

select          id

               ,1.0000

            *   last_value (field1) over
                (
                    partition by    id
                    order by        field3
                    rows between    unbounded preceding 
                                and unbounded following
                )

            /   nullifzero (field1)

            -   1   as fieldnew

from        mytable

where       field3 = td_month_begin (current_date) - interval '2' month
         or field3 = td_month_begin (current_date) - interval '1' month 

qualify     field3 = min (field3) over (partition by id)
;

或

select          id

               ,1.0000

            *   min (field1) over
                (
                    partition by    id
                    order by        field3
                    rows between    1 following
                                and 1 following
                )

            /   nullifzero (field1)

            -   1   as fieldnew

where       field3 = td_month_begin (current_date) - interval '2' month
         or field3 = td_month_begin (current_date) - interval '1' month 

from        mytable

qualify     field3 = min (field3) over (partition by id)
;

+----+----------+
| Id | fieldnew |
+----+----------+
| 1  | 0.7500   |
+----+----------+
| 2  | 0.5000   |
+----+----------+
| 5  | 0.7368   |
+----+----------+

在teradata sql 中划分两条记录并打印一条记录

Divide two records in teradata sql and print one record

sql

teradata

甲骨文

天睿