问题
I have the following data, composed of the A value, ordered by MM (month).
The B column is computed as GREATEST(current value of A + previous value of B, 0) in a spreadsheet-like fashion.
How can I compute B using a SQL Query?
- I tried using Analytic Functions, but I was unable to succeed.
- I know there is the Model Clause; I found a similar example, but I don't know where to begin.
I am using Oracle 10g, therefore I cannot use recursive queries.
Here is my test data:
MM | A | B
-----------+--------+------
2012-01-01 | 800 | 800
2012-02-01 | 1900 | 2700
2012-03-01 | 1750 | 4450
2012-04-01 | -20000 | 0
2012-05-01 | 900 | 900
2012-06-01 | 3900 | 4800
2012-07-01 | -2600 | 2200
2012-08-01 | -2600 | 0
2012-09-01 | 2100 | 2100
2012-10-01 | -2400 | 0
2012-11-01 | 1100 | 1100
2012-12-01 | 1300 | 2400
And here is the "table definition":
select t.* from (
select date'2012-01-01' as mm, 800 as a from dual union all
select date'2012-02-01' as mm, 1900 as a from dual union all
select date'2012-03-01' as mm, 1750 as a from dual union all
select date'2012-04-01' as mm, -20000 as a from dual union all
select date'2012-05-01' as mm, 900 as a from dual union all
select date'2012-06-01' as mm, 3900 as a from dual union all
select date'2012-07-01' as mm, -2600 as a from dual union all
select date'2012-08-01' as mm, -2600 as a from dual union all
select date'2012-09-01' as mm, 2100 as a from dual union all
select date'2012-10-01' as mm, -2400 as a from dual union all
select date'2012-11-01' as mm, 1100 as a from dual union all
select date'2012-12-01' as mm, 1300 as a from dual
) t;
回答1:
So let's unleash the MODEL clause (a device whose mystery is only exceeded by its power) on this problem:
with data as (
select date'2012-01-01' as mm, 800 as a from dual union all
select date'2012-02-01' as mm, 1900 as a from dual union all
select date'2012-03-01' as mm, 1750 as a from dual union all
select date'2012-04-01' as mm, -20000 as a from dual union all
select date'2012-05-01' as mm, 900 as a from dual union all
select date'2012-06-01' as mm, 3900 as a from dual union all
select date'2012-07-01' as mm, -2600 as a from dual union all
select date'2012-08-01' as mm, -2600 as a from dual union all
select date'2012-09-01' as mm, 2100 as a from dual union all
select date'2012-10-01' as mm, -2400 as a from dual union all
select date'2012-11-01' as mm, 1100 as a from dual union all
select date'2012-12-01' as mm, 1300 as a from dual
)
select mm, a, b
from (
-- Add a dummy value for b, making it available to the MODEL clause
select mm, a, 0 b
from data
)
-- Generate a ROW_NUMBER() dimension, in order to access rows by RN
model dimension by (row_number() over (order by mm) rn)
-- Spreadsheet values / measures involved in calculations are mm, a, b
measures (mm, a, b)
-- A single rule will do. Any value of B should be calculated according to
-- GREATEST([previous value of B] + [current value of A], 0)
rules (
b[any] = greatest(nvl(b[cv(rn) - 1], 0) + a[cv(rn)], 0)
)
The above yields:
MM A B
01.01.2012 800 800
01.02.2012 1900 2700
01.03.2012 1750 4450
01.04.2012 -20000 0
01.05.2012 900 900
01.06.2012 3900 4800
01.07.2012 -2600 2200
01.08.2012 -2600 0
01.09.2012 2100 2100
01.10.2012 -2400 0
01.11.2012 1100 1100
01.12.2012 1300 2400
回答2:
I came up with a user-defined aggregate function
create or replace type tsum1 as object
(
total number,
static function ODCIAggregateInitialize(nctx IN OUT tsum1 )
return number,
member function ODCIAggregateIterate(self IN OUT tsum1 ,
value IN number )
return number,
member function ODCIAggregateTerminate(self IN tsum1,
retVal OUT number,
flags IN number)
return number,
member function ODCIAggregateMerge(self IN OUT tsum1,
ctx2 IN tsum1)
return number
)
/
create or replace type body tsum1
is
static function ODCIAggregateInitialize(nctx IN OUT tsum1)
return number
is
begin
nctx := tsum1(0);
return ODCIConst.Success;
end;
member function ODCIAggregateIterate(self IN OUT tsum1,
value IN number )
return number
is
begin
self.total := self.total + value;
if (self.total < 0) then
self.total := 0;
end if;
return ODCIConst.Success;
end;
member function ODCIAggregateTerminate(self IN tsum1,
retVal OUT number,
flags IN number)
return number
is
begin
retVal := self.total;
return ODCIConst.Success;
end;
member function ODCIAggregateMerge(self IN OUT tsum1,
ctx2 IN tsum1)
return number
is
begin
self.total := self.total + ctx2.total;
return ODCIConst.Success;
end;
end;
/
CREATE OR REPLACE FUNCTION sum1(input number)
RETURN number
PARALLEL_ENABLE AGGREGATE USING tsum1;
/
Here is the query
with T1 as(
select date'2012-01-01' as mm, 800 as a from dual union all
select date'2012-02-01' as mm, 1900 as a from dual union all
select date'2012-03-01' as mm, 1750 as a from dual union all
select date'2012-04-01' as mm, -20000 as a from dual union all
select date'2012-05-01' as mm, 900 as a from dual union all
select date'2012-06-01' as mm, 3900 as a from dual union all
select date'2012-07-01' as mm, -2600 as a from dual union all
select date'2012-08-01' as mm, -2600 as a from dual union all
select date'2012-09-01' as mm, 2100 as a from dual union all
select date'2012-10-01' as mm, -2400 as a from dual union all
select date'2012-11-01' as mm, 1100 as a from dual union all
select date'2012-12-01' as mm, 1300 as a from dual
)
select mm
, a
, sum1(a) over(order by mm) as b
from t1
Mm a b
----------------------------
01.01.2012 800 800
01.02.2012 1900 2700
01.03.2012 1750 4450
01.04.2012 -20000 0
01.05.2012 900 900
01.06.2012 3900 4800
01.07.2012 -2600 2200
01.08.2012 -2600 0
01.09.2012 2100 2100
01.10.2012 -2400 0
01.11.2012 1100 1100
01.12.2012 1300 2400
回答3:
with sample_data as (
select date'2012-01-01' as mm, 800 as a from dual union all
select date'2012-02-01' as mm, 1900 as a from dual union all
select date'2012-03-01' as mm, 1750 as a from dual union all
select date'2012-04-01' as mm, -20000 as a from dual union all
select date'2012-05-01' as mm, 900 as a from dual union all
select date'2012-06-01' as mm, 3900 as a from dual union all
select date'2012-07-01' as mm, -2600 as a from dual union all
select date'2012-08-01' as mm, -2600 as a from dual union all
select date'2012-09-01' as mm, 2100 as a from dual union all
select date'2012-10-01' as mm, -2400 as a from dual union all
select date'2012-11-01' as mm, 1100 as a from dual union all
select date'2012-12-01' as mm, 1300 as a from dual
)
select mm,
a,
greatest(nvl(a,0) + lag(a,1,0) over (order by mm), 0) as b
from sample_data;
It does however not produce this line:
2012-05-01 | 900 | 900
because it calculates 900 - 20000 in that row, and zero is bigger than the result of that. You can "fix" that if you use the abs function to get rid of the negative value in the computation.
回答4:
Sorry if this is off topic, given the Oracle version of the question, but we can now use the SQL:2016 MATCH_RECOGNIZE clause:
select * from t
match_recognize(
order by mm
measures case classifier() when 'POS' then sum(a) else 0 end as b
all rows per match
pattern (pos* neg{0,1})
define pos as sum(a) > 0
);
来源：https://stackoverflow.com/questions/12243488/oracle-sql-analytic-query-recursive-spreadsheet-like-running-total