How can I find the first row where the negative value starts in Oracle? Below is an example.
77
74
67
56
42
24
20
19
-17
-28
-31
-36
I would like to read the row -17 and do some operations on that row.
Any help is greatly appreciated. Thanks
If you want the complete row, you can use:
select t.*
from t
where n < 0
order by n asc
fetch first 1 row only;
If you have another column that specifies the ordering of the rows, then:
select t.*
from t
where n < 0
order by <ordering col> asc
fetch first 1 row only;
To select the maximum negative number you can do:
select max(n)
from t
where n < 0
Remember that in relational database tables, rows do not have inherent ordering. Therefore, in the absence of an ordering criteria, there's no such a thing as "first row where the negative value starts".
Assuming you have a column for sorting, which defines the order of the rows, it could look like this:
with t as (
select 77 a, 1 row_order from dual union all
select 74 a, 2 row_order from dual union all
select 67 a, 3 row_order from dual union all
select 56 a, 4 row_order from dual union all
select 42 a, 5 row_order from dual union all
select 24 a, 6 row_order from dual union all
select 20 a, 7 row_order from dual union all
select 19 a, 8 row_order from dual union all
select -17 a, 9 row_order from dual union all
select -28 a, 10 row_order from dual union all
select -31 a, 11 row_order from dual union all
select -36 a, 12 row_order from dual
), t1 as (
select a, row_number() over (partition by case when a < 0 then 0 else 1 end order by row_order) rn from t
)
select * from t1 where rn = 1 and a < 0;
It's using a window function in order to determine the first rows (here for positive a's and negative a's)
Then it selects the first row encountered that is negative.
Related
I have the below table:
LAUFD
ID
NEXDT
ORDER_ROW
20140305
C1
20140310
14
20140226
C1
20140305
13
20131125
C1
20131126
12
20131021
C1
20131022
11
20130821
C1
20130828
10
20130814
C1
20130821
9
20130807
C1
20130814
8
20130731
C1
20130807
7
20130724
C1
20130731
6
20130710
C1
20130724
5
20130708
C1
20130709
4
20130624
C1
20130707
3
20130603
C1
20130608
2
20130527
C1
20130603
1
I would like to have the below output:
ID
START
END
C1
20140226
20140310
The logic is: if, ordering ID by order_row, the field NEXDT is equal or equal+1 or equal+2 to the field LAUFD of the next order_row, then continue with the next entry. If not, generate an entry in the output table with the start (earliest LAUFD) and end (latest NEXDT).
Basically, it's the same question as in Oracle SQL row concatenation by periods but I'd like just the latest period as an output.
Looks like this is what you need:
with t (LAUFD, ID, NEXDT, ORDER_ROW) as (
select 20140305,'C1', 20140310, 14 from dual union all
select 20140226,'C1', 20140305, 13 from dual union all
select 20131125,'C1', 20131126, 12 from dual union all
select 20131021,'C1', 20131022, 11 from dual union all
select 20130821,'C1', 20130828, 10 from dual union all
select 20130814,'C1', 20130821, 9 from dual union all
select 20130807,'C1', 20130814, 8 from dual union all
select 20130731,'C1', 20130807, 7 from dual union all
select 20130724,'C1', 20130731, 6 from dual union all
select 20130710,'C1', 20130724, 5 from dual union all
select 20130708,'C1', 20130709, 4 from dual union all
select 20130624,'C1', 20130707, 3 from dual union all
select 20130603,'C1', 20130608, 2 from dual union all
select 20130527,'C1', 20130603, 1 from dual
)
,t1 as (select id, order_row, to_date(laufd,'yyyymmdd') as laufd_dt, to_date(nexdt,'yyyymmdd') as nexdt_dt from t)
select *
from t1
match_recognize (
partition by id
order by order_row desc
measures
min(x.laufd_dt) as dt_start,
max(a.nexdt_dt) as dt_end,
x.laufd_dt-next(x.nexdt_dt) as dates_diff
one row per match
pattern(a x+ y* z*)
define
x as x.order_row=prev(order_row)-1 and prev(laufd_dt)-nexdt_dt<=3
,y as x.order_row=prev(order_row)-1
);
For just the latest period, you could use the previous solution. But instead, look for the first "break". Then only use the rows since that break;
select id, min(laufd), max(nextdt),
row_number() over (partition by id order by min(laufd)) as period
from (select t.*,
sum(case when prev_nextdt >= laufd - interval '2' day then 0 else 1 end) over
(partition by id order by order_row range desc) as grp,
sum(case when prev_nextdt >= laufd - interval '2' day then 0 else 1 end) over (partition by id) as num_grps
from (select t.id, t.order_row, -- any other columns you need
to_date(laufd, 'YYYYMMDD') as laufd,
to_date(nextdt, 'YYYYMMDD') as next_dt,
lag(to_date(nextdt, 'YYYYMMDD')) over (partition by id order by order_row) as prev_nextdt
from t
) t
) t
where num_grps = grp
group by id;
This is basically the same logic. It just keeps the first group.
I have the following data (order of records as in the example):
A B
1 10
1 20
1 30
1 40
2 50
2 65
2 75
1 89
1 100
from SQL:
with x as (
select A, B
from (
select 1 as A, 10 as B from dual
union all
select 1 as A, 20 as B from dual
union all
select 1 as A, 30 as B from dual
union all
select 1 as A, 40 as B from dual
union all
select 2 as A, 50 as B from dual
union all
select 2 as A, 65 as B from dual
union all
select 2 as A, 75 as B from dual
union all
select 1 as A, 89 as B from dual
union all
select 1 as A, 100 as B from dual
)
)
select A, B
from X
I want to group the data for each change of value in column A,
I want to get the following result:
A MIN(B) MAX(B)
1 10 40
2 50 75
1 89 100
How to get such a result in the ORACLE 11. I would expect a simple implementation...
This is a gaps and islands problem, solved using row_number analytic function
SELECT a,
MIN(b),
MAX(b)
FROM (
SELECT x.*,
ROW_NUMBER() OVER(
ORDER BY b
) - ROW_NUMBER() OVER(
PARTITION BY a
ORDER BY b
) AS seq
FROM x
)
GROUP BY a,
seq;
Demo
I want to achieve a running value, but condition is reset on some specific column value.
Here is my select statement:
with tbl(emp,salary,ord) as
(
select 'A',1000,1 from dual union all
select 'B',1000,2 from dual union all
select 'K',1000,3 from dual union all
select 'A',1000,4 from dual union all
select 'B',1000,5 from dual union all
select 'D',1000,6 from dual union all
select 'B',1000,7 from dual
)
select * from tbl
I want to reset count on emp B if the column value is B, then count is reset to 0 and started again increment by 1:
emp salary ord running_count
A 1000 1 0
B 1000 2 1
K 1000 3 0
A 1000 4 1
B 1000 5 2
D 1000 6 0
B 1000 7 1
Here order column is ord.
I want to achieve the whole thing by select statement, not using the cursor.
You want to define groups were the counting takes place. Within a group, the solution is row_number().
You can define the group by doing a cumulative sum of B values. Because B ends the group, you want to count the number of B after each record.
This results in:
select t.*,
row_number() over (partition by grp order by ord) - 1 as running_count
from (select t.*,
sum(case when emp = 'B' then 1 else 0 end) over (order by ord desc) as grp
from tbl t
) t;
select distinct account_num from account order by account_num;
The above query gave the below result
account_num
1
2
4
7
12
18
24
37
45
59
I want to split the account_num column into tuple of three account_num's like (1,2,4);(7,12,18);(24,37,45),(59); The last tuple has only one entry as there are no more account_num's left. Now I want a query to output the min and max of each tuple. (please observe that the max of one tuple is less than the min of the next tuple). Output desired is shown below
1 4
7 18
24 45
59 59
Edit: I have explained my requirement in the best way I could
You can use the example below as a scratch, this is only based on information you have provided so far. For further documentation, you can consult Oracle's analytical functions docs:
with src as( --create a source data
select 1 col from dual union
select 2 from dual union
select 4 from dual union
select 7 from dual union
select 12 from dual union
select 18 from dual union
select 24 from dual union
select 37 from dual union
select 45 from dual union
select 59 from dual
)
select
col,
decode(col_2, 0, max_col, col_2) col_2 -- for the last row we get the maximum value for the row
from (
select
col,
lead(col, 2, 0) over (order by col) col_2, -- we get the values from from two rows behind
max(col) over () max_col, -- we get the max value to be used for the last row in the result
rownum rn from src -- we get the rownum to handle the final output
) where mod(rn - 1, 3) = 0 -- only get rows having a step of two
This is another solution.
SELECT *
FROM (SELECT DISTINCT MIN(val) over(PARTITION BY gr) min_,
MAX(val) over(PARTITION BY gr) max_
FROM (SELECT val,
decode(trunc(rn / 3), rn / 3, rn / 3, ceil(rn / 3)) gr
FROM (SELECT val,
row_number() over(ORDER BY val) rn
FROM (select distinct account_num from account order by account_num)))) ORDER BY min_
UPDATED
Solution without analytic function.
SELECT MIN(val) min_,
MAX(val) max_
FROM (SELECT val,
ceil(rn / 3) gr
FROM (SELECT val,
rownum rn
FROM A_DEL_ME)) GROUP BY gr
Please add more information on what you want to do. What is the connection between account_number 1 and number 4, 7 and 18? Is there any? If not, why would you want to split this into two columns and what is the rule for splitting it?
With what you have posted, you could do something like this:
select 1 as account_num, 4 as account_num1 from dual
union all select 7 as account_num, 18 as account_num1 from dual
...
and so on, but I don't see the use for this.
In my table I want to see a list of unoccupied id numbers in a certain range.
For example there are 10 records in my table with id's: "2,3,4,5,10,12,16,18,21,22" and say that I want to see available ones between 1 and 25. So I want to see a list like:
1,6,7,89,11,13,14,15,17,19,20,23,24,25
How should I write my sql query?
Select the numbers form 1 to 25 and show only those that are not in your table
select n from
( select rownum n from dual connect by level <= 25)
where n not in (select id from table);
Let's say you a #numbers table with three numbers -
CREATE TABLE #numbers (num INT)
INSERT INTO #numbers (num)
SELECT 1
UNION
SELECT 3
UNION
SELECT 6
Now, you can use CTE to generate numbers recursively from 1-25 and deselect those which are in your #numbers table in the WHERE clause -
;WITH n(n) AS
(
SELECT 1
UNION ALL
SELECT n+1 FROM n WHERE n < 25
)
SELECT n FROM n
WHERE n NOT IN (select num from #numbers)
ORDER BY n
OPTION (MAXRECURSION 25);
You can try using the "NOT IN" clause:
select
u1.user_id + 1 as start
from users as u1
left outer join users as u2 on u1.user_id + 1 = u2.id
where
u2.id is null
see also SQL query to find Missing sequence numbers
You need LISTAGG to get the output in a single row.
SQL> WITH DATA1 AS(
2 SELECT LEVEL rn FROM dual CONNECT BY LEVEL <=25
3 ),
4 data2 AS(
5 SELECT 2 num FROM dual UNION ALL
6 SELECT 3 FROM dual UNION ALL
7 SELECT 4 from dual union all
8 SELECT 5 FROM dual UNION ALL
9 SELECT 10 FROM dual UNION ALL
10 SELECT 12 from dual union all
11 SELECT 16 from dual union all
12 SELECT 18 FROM dual UNION ALL
13 SELECT 21 FROM dual UNION ALL
14 SELECT 22 FROM dual)
15 SELECT listagg(rn, ',')
16 WITHIN GROUP (ORDER BY rn) num_list FROM data1
17 WHERE rn NOT IN(SELECT num FROM data2)
18 /
NUM_LIST
----------------------------------------------------
1,6,7,8,9,11,13,14,15,17,19,20,23,24,25
SQL>