I have two requests that do the same thing, but for different groups.
SELECT count() FROM Table1 WHERE label = 'label1' AND group IN (0) AND date >= '2018-01-01' AND date <= '2018-02-01';
SELECT count() FROM Table1 WHERE label = 'label1' AND group IN (3, 4, 5, 6, 7, 8) AND date >= '2018-03-01' AND date <= '2018-03-01';
Can I split it one query?
You can use..
UNION ALL
..between the two queries to only get one result returned. It will still be two queries however, if you want it to be one query, you can use CASE WHEN .
SELECT
SUM(CASE WHEN group = 0 AND date BETWEEN '2018-01-01' AND '2018-02-01' THEN 1 END) As Group1,
SUM(CASE WHEN group IN (3, 4, 5, 6, 7, 8) AND date BETWEEN '2018-03-01' AND '2018-03-01 THEN 1 END) As Group2
FROM Table1
WHERE label = 'label1'
Related
I have this
RowNum
Value
1
X
2
X
3
Y
4
Z
5
Z
6
Z
7
V
and I want something like this
RowNum
Value
1
X
1
X
2
Y
3
Z
3
Z
3
Z
4
V
How can I do that in Oracle?
Thanks
Here is one way - using the match_recognize clause, available since Oracle 12.1. Note that rownum is a reserved keyword, so it can't be a column name; I changed it to rnum in the input and rn in the output, adapt as needed.
The with clause is not part of the query - it's there just to include your sample data. Remove it before using the query on your actual data.
with
inputs (rnum, value) as (
select 1, 'X' from dual union all
select 2, 'X' from dual union all
select 3, 'Y' from dual union all
select 4, 'Z' from dual union all
select 5, 'Z' from dual union all
select 6, 'Z' from dual union all
select 7, 'V' from dual
)
select rn, value
from inputs
match_recognize (
order by rnum
measures match_number() as rn
all rows per match
pattern ( a+ )
define a as value = first(value)
);
RN VALUE
-- -----
1 X
1 X
2 Y
3 Z
3 Z
3 Z
4 V
In Oracle 11.2 and earlier, you can use the start-of-group method (the flags created in the subquery and counted in the outer query):
select count(flag) over (order by rnum) as rn, value
from (
select rnum, value,
case lag(value) over (order by rnum)
when value then null else 1 end as flag
from inputs
)
;
Consider following table where I am doing row data multiplication:
with v1 (member_id, the_number) as
(
select 1, 3 from dual union all
select 1, 5 from dual union all
select 2, 2 from dual union all
select 2, 3 from dual union all
select 2, 4 from dual union all
select 3, 9 from dual union all
select 3, 3 from dual union all
select 3, 2 from dual
)
select member_id, EXP(SUM(LN(the_number))) from v1
GROUP BY member_id;
It gives the correct result as:
MEMBER_ID EXP(SUM(LN(THE_NUMBER)))
1 15
2 24
3 54
The moment I put a negative value in the the_number column, I get the following Oracle error: ORA-01428: argument 'x' is out of range This is because the range for LN () argument is > 0.
How can I modify the query so that I can have negative values as well in the_number column? I am using Oracle 11g.
Get the product of the absolute values of the numbers and finally multiply by -1 or 1 depending on whether there is an odd or even number of negative numbers:
select
member_id,
CASE WHEN MOD(SUM(CASE WHEN the_number < 0 THEN 1 ELSE 0 END), 2) = 1 THEN -1 ELSE 1 END *
EXP(SUM(LN(ABS(the_number)))) from v1
GROUP BY member_id;
See the demo.
Column 'amount' has value 5 in first row, and value -10 in second row.
Is there a way to make oracle's sum(amount) over() function to return 0 instead of -5 for the second row?
Blatantly using Rajesh Chamarthi's example source: but altering to show more negative and positive... and showing how a case would change all the negative to zero while maintaining the other amounts...
with t as (
select 5 as x, 1 as id from dual
union all
select -10, 2 as id from dual
union all
select 7, 3 as id from dual
union all
select -5, 4 as id from dual
union all
select -2, 5 as id from dual
),
B as (select t.x,
case when sum(x) over (order by id) < 0 then 0
else sum(x) over (order by id)
end Amount
from t
order by id)
Select X, Case when amount < 0 then 0 else amount end as Amount from B;
T Amount
5 5
-10 0
7 2
-5 0
-2 0
----Attempt 2 (1st attempt preserved as comments below reference it)
I couldn't figure out how to interrupt the window function to reset the value to 0 when amount fell below 0... so I used a recursive CTE which gave me greater control.
If id's are not sequential, we could add a row_Number so we have an ID to join on... or we could use min() where > oldID. I assumed we have a single key unique ID or some way of "Sorting" the records in the order you want the sum to occur...
with aRaw as (
select 5 as x, 15 as id from dual
union all
select -10, 20 as id from dual
union all
select 7, 32 as id from dual
union all
select 2, 46 as id from dual
union all
select -15, 55 as id from dual
union all
select 3, 68 as id from dual
),
t as (Select A.*, Row_number() over (order by ID) rn from aRAW A),
CTE(RN, ID, x, SumX) AS (
Select T.RN, T.ID, x, X from t WHERE ID = (Select min(ID) from t)
UNION ALL
Select T.RN, T.ID, T.X, case when T.X+CTE.SumX < 0 then 0 else T.X+Cte.sumX end from T
INNER JOIN CTE
on CTE.RN+1=T.RN)
Select * from cte;
.
CTE: ARaw is just a sample data set
CTE: T adds a sequental row number incase there are gaps in the IDs allowing for a more simple joining approach on the recursive CTE.
CTE: CTE is the recursive CTE that keeps a running total and has a case statement to reset the running total to 0 when it falls below 0
You could use a case statement, but that would not be a true running total
with t as (
select 5 as x, 1 as id from dual
union all
select -10, 2 as id from dual
union all
select 20, 3 as id from dual
union all
select 30, 4 as id from dual
union all
select 10, 5 as id from dual
)
select t.x,
case when sum(x) over (order by id) < 0 then 0
else sum(x) over (order by id)
end running_total
from t
order by id;
X RUNNING_TOTAL
5 5
-10 0
20 15
30 45
10 55
---------------
MASTER TABLE
---------------
DATA_KEY NUMBER
TEXT VARCHAR2(2000)
ORDER_NO NUMBER
---------------
DETAIL TABLE
---------------
DATA_KEY NUMBER
SIMILAR_DATA_KEY NUMBER
DISTANCE_COUNT NUMBER
---------------
INSERT QUERY
---------------
INSERT INTO DETAIL
(
SELECT DATA_KEY, SIMILAR_DATA_KEY, DISTANCE_COUNT
FROM
(
SELECT A.DATA_KEY AS DATA_KEY, B.DATA_KEY AS SIMILAR_DATA_KEY,
UTL_MATCH.EDIT_DISTANCE(A.TEXT, B.TEXT) AS DISTANCE_COUNT
FROM
(SELECT DATA_KEY, TEXT, ORDER_NO FROM MASTER) A
INNER JOIN
(SELECT DATA_KEY, TEXT, ORDER_NO FROM MASTER) B
ON (A.ORDER_NO < B.ORDER_NO)
)
WHERE DISTANCE_COUNT <= 5
)
I need compare MASTER table TEXT field with other TEXT field.
indexes are not exist.
master table 90,000 rows.
ORDER_NO field is for avoid duplicated compare. (1 .. 90000)
=============================================================
A.ORDER_NO < B.ORDER_NO
------------------------------------------
1, 1 <- exclude
1, 2 <- join
1, 3 <- join
1, 4 <- join
..
2, 1 <- exclude
2, 2 <- exclude
2, 3 <- join
2, 4 <- join
...
3, 1 <- exclude
3, 2 <- exclude
3, 3 <- exclude
3, 4 <- join
1. NOT need compare 1 and 1
2. Need compare 1 and 2
3. NOT need compare 2 and 1 (because, duplicate 2.)
so, for decrease compare count...
=============================================================
Slow zone is (WHERE DISTANCE_COUNT <= 5) ?
Slow zone is comparing rows (90000*89999/2) ?
Query elapse time is 7 days.
6,000 rows inserted to DETAIL table.
How to speed up?
I'm sorry for poor English...
I have this query
SELECT code, username, week1money, week2money, week3money, week4money FROM(
--subquery goes here
)
How to select the top two weeks, i.e. weeks with the highest value? I want to sum the top two weeks to be precise.
If I understand correct you want to get 2 top values per every (code, username) row and (code, username) is a key of recordset.
Supposing you can have two top weeks with the same values and you don't have nulls this might be one of solutions:
SQL> with t (id, code, week1, week2, week3, week4)
2 as (
3 select 1, 'a', 10, 15, 11, 8 from dual union all
4 select 2, 'b', 7, 4, 2, 9 from dual union all
5 select 3, 'c', 3, 3, 1, 0 from dual
6 )
7 select id, code, max(week) first_top, min(week) next_top from (
8 select id, code, row_number() over(partition by id, code order by week desc) rnk, week
9 from (
10 select t.id, t.code,
11 decode(r.rn,1,week1,2,week2,3,week3,4,week4) week
12 from t,
13 (select rownum rn from dual connect by level <= 4) r
14 ))
15 where rnk in (1,2)
16 group by id, code
17 /
ID C FIRST_TOP NEXT_TOP
---------- - ---------- ----------
3 c 3 3
1 a 15 11
2 b 9 7
If you have non-null and different values in weeks you can use something like:
SQL> with t (id, code, week1, week2, week3, week4)
2 as (
3 select 1, 'a', 10, 15, 11, 8 from dual union all
4 select 2, 'b', 7, 4, 2, 9 from dual union all
5 select 3, 'c', 3, 2, 1, 0 from dual
6 )
7 select id, code
8 , greatest(week1, week2, week3, week4) first_top
9 , greatest(
10 case when week1 < greatest(week1, week2, week3, week4) then week1 else -1e28 end,
11 case when week2 < greatest(week1, week2, week3, week4) then week2 else -1e28 end,
12 case when week3 < greatest(week1, week2, week3, week4) then week3 else -1e28 end,
13 case when week4 < greatest(week1, week2, week3, week4) then week4 else -1e28 end
14 ) second_top
15 from t
16 /
ID C FIRST_TOP SECOND_TOP
---------- - ---------- ----------
1 a 15 11
2 b 9 7
3 c 3 2
But to get the right solution more details are required.
Answering my question...
select * from(
select * from(
select week1money col from dual
union
select week2money col from dual
union
select week3money col from dual
union
select week4money col from dual
) order by col desc
) where rownum < 3
Using GREATESTS() also may help.