I have a table t1 and there is a column named days, so I want to group the days by 1 to 5 days and 6-15 days and more than 15 days then calculate the count for each group, but I don't know how to write the sql, can anyone tell me? the result should be looked like below:
Number Scope(days)
5 1-5
7 6-15
10 15+
If I understand well, this could be a way:
with t1(days) as (
select 1 from dual union all
select 2 from dual union all
select 5 from dual union all
select 6 from dual union all
select 7 from dual union all
select 8 from dual union all
select 10 from dual union all
select 11 from dual union all
select 16 from dual union all
select 17 from dual union all
select 19 from dual union all
select 19 from dual
)
/* the query */
select count(*),
case
when days between 1 and 5 then '1-5'
when days between 6 and 15 then '6-15'
else '+15'
end
from t1
group by case
when days between 1 and 5 then '1-5'
when days between 6 and 15 then '6-15'
else '+15'
end
that gives:
COUNT(*) CASE
---------- ----
3 1-5
4 +15
5 6-15
The idea is to aggregate by something that say the "group" where every number is, and you can easily build such an information with a CASE.
Accordin to Jarlh's suggestion, this can be re-written as
select count(*), the_group
from (
select case
when days between 1 and 5 then '1-5'
when days between 6 and 15 then '6-15'
else '+15'
end the_group
from t1
)
group by the_group
This obviously assumes that you only have positive numbers.
Related
I have this query that needs to be executed for oracle sql instead of mysql which is where it originally came from, but I have the ADDDATE() function which I don't see any other alternative than DateAdd since it needs more parameters than I really need..
Apart from that, if I try to execute it, it also indicates an error in the
SELECT 0 i UNION.................
part, saying the following ORA-00923: FROM keyword not found where expected
Maybe in oracle it is not allowed to do a select 0 union select 1 union...
Any suggestions or help I appreciate it, thanks
SELECT
ADDDATE('1970-01-01', t4.i * 10000 + t3.i * 1000 + t2.i * 100 + t1.i * 10 + t0.i) selected_date
FROM
(
SELECT 0 i UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5 UNION SELECT 6 UNION SELECT 7 UNION SELECT 8 UNION SELECT 9
) t0,
(
SELECT 0 i UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5 UNION SELECT 6 UNION SELECT 7 UNION SELECT 8 UNION SELECT 9
) t1,
(
SELECT 0 i UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5 UNION SELECT 6 UNION SELECT 7 UNION SELECT 8 UNION SELECT 9
) t2,
(
SELECT 0 i UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5 UNION SELECT 6 UNION SELECT 7 UNION SELECT 8 UNION SELECT 9
) t3,
(
SELECT 0 i UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5 UNION SELECT 6 UNION SELECT 7 UNION SELECT 8 UNION SELECT 9
) t4
In Oracle you must select from the one-row table dual in order to select one row. You cannot select without a from clause.
If you want to generate dates, you'll write a standard SQL recursive CTE. (And this is the typical approach now in MySQL, too, since version 8.0.)
Here is an example selecting all days for 1970:
with dates (dt) as
(
select date '1970-01-01' from dual
union all
select dt + interval '1' day from dates where dt < date '1970-12-31'
)
select dt from dates;
Here is another way to SELECT a list of dates for the year 1970. Adjust the starting and ending dates if you want different years or the INTERVAL if you want different periods like seconds, minutes, hours…
ALTER SESSION SET NLS_DATE_FORMAT = 'DD-MON-YYYY HH24:MI:SS';
with dt (dt, interv) as (
select date '1970-01-01', numtodsinterval(1,'DAY') from dual
union all
select dt.dt + interv, interv from dt
where dt.dt + interv <= date '1970-12-31')
select dt from dt;
/
I have Oracle SQL Database with values like:
Date Number
20.04.22 1
20.04.22 2
20.04.22 3
20.04.22 4
20.04.22 5
20.04.22 6
21.04.22 1
21.04.22 2
21.04.22 3
21.04.22 4
21.04.22 5
21.04.22 6
Now, I want to select 3 rows, starting on number==value. The value is coming from somewhere else. For example, that value is 2, then I want to have 20.04.22 number 2-5.
How do I write a query if I want to have the numbers between two days? Like when the value is 6, then I need to have 20.04.22 number 6 and 21.04.22 number 1-2...
If its just one day, I can use where number >= [number i choose]
Thanks!
From Oracle 12, you can use:
SELECT "DATE", "NUMBER"
FROM (
SELECT t.*,
COUNT(CASE "NUMBER" WHEN 6 THEN 1 END) OVER (ORDER BY "DATE", "NUMBER")
AS cnt
FROM table_name t
ORDER BY "DATE", "NUMBER"
)
WHERE cnt > 0
FETCH FIRST 3 ROWS ONLY;
Before Oracle 12, you can use:
SELECT "DATE", "NUMBER"
FROM (
SELECT t.*,
COUNT(CASE "NUMBER" WHEN 6 THEN 1 END) OVER (ORDER BY "DATE", "NUMBER")
AS cnt
FROM table_name t
ORDER BY "DATE", "NUMBER"
)
WHERE cnt > 0
AND ROWNUM <= 3;
(Note: DATE and NUMBER are reserved words and it is considered bad practice to use them as identifiers and, if you do, then must be quoted identifiers.)
Which, for your sample data:
CREATE TABLE table_name ("DATE", "NUMBER") AS
SELECT DATE '2022-04-20', 1 FROM DUAL UNION ALL
SELECT DATE '2022-04-20', 2 FROM DUAL UNION ALL
SELECT DATE '2022-04-20', 3 FROM DUAL UNION ALL
SELECT DATE '2022-04-20', 4 FROM DUAL UNION ALL
SELECT DATE '2022-04-20', 5 FROM DUAL UNION ALL
SELECT DATE '2022-04-20', 6 FROM DUAL UNION ALL
SELECT DATE '2022-04-21', 1 FROM DUAL UNION ALL
SELECT DATE '2022-04-21', 2 FROM DUAL UNION ALL
SELECT DATE '2022-04-21', 3 FROM DUAL UNION ALL
SELECT DATE '2022-04-21', 4 FROM DUAL UNION ALL
SELECT DATE '2022-04-21', 5 FROM DUAL UNION ALL
SELECT DATE '2022-04-21', 6 FROM DUAL;
Outputs:
DATE
NUMBER
20-APR-22
6
21-APR-22
1
21-APR-22
2
db<>fiddle here
let’s imagine here is my report:
select 1 from dual
union all
select 2 from dual
union all
select 3 from dual
union all
select 4 from dual
Output:
1
2
3
4
If e.g. i want to hide part of the report in an easy way I simply add 1 = 2 to where clause:
select 1 from dual
union all
select 2 from dual
union all
select 3 from dual where 1 = 2
union all
select 4 from dual
Output:
1
2
4
Cool! However, when select uses aggregate functions, this trick does not help. The row is generated anyway:
select 1 from dual
union all
select 2 from dual
union all
select max(3) from dual where 1 = 2
union all
select 4 from dual
Output:
1
2
(null)
4
Someone maybe know a simple, easy way to hide such rows?
An aggregation query with no group by always returns exactly one row -- even if no rows are being aggregated.
You want to filter after the aggregation, so use having:
select max(3) from dual having 1 = 2
I've a table (T1) with one column (C1) with the below values
1
2
3
5
6
8
9
10
I want the output to print the continuous sequences with start and ending points like below.
1-3
5-6
8-10
Could you please help?
Any Database is fine.
Oracle: sample data first, while code you really need begins at line #11.
SQL> with t1 (c1) as
2 (select 1 from dual union all
3 select 2 from dual union all
4 select 3 from dual union all
5 select 5 from dual union all
6 select 6 from dual union all
7 select 8 from dual union all
8 select 9 from dual union all
9 select 10 from dual
10 )
11 select min(c1), max(c1)
12 from (select c1, c1 - row_number() over (order by c1) rn
13 from t1
14 )
15 group by rn
16 order by rn;
MIN(C1) MAX(C1)
---------- ----------
1 3
5 6
8 10
SQL>
You can use the following query. I have tested it with SQL Server, but I think it will work without modifications in Oracle:
create table t1(c1 int);
insert into t1
select *
from (values(1),(2),(3),(5),(6),(8),(9),(10))t(x);
select case when count(*) >1 then
concat(min(c1),'-',max(c1))
else concat(max(c1),'')
end as concat_cs
from (
select c1
,ROW_NUMBER() over(order by c1 asc) as rnk
,c1 - ROW_NUMBER() over(order by c1 asc) as grp
from t1
)x
group by x.grp
Output
concat_cs
1-3
5-6
8-10
with stab as (
select 1 as val from dual union all
select 2 as val from dual union all
select 3 as val from dual union all
select 5 as val from dual union all
select 6 as val from dual union all
select 8 as val from dual union all
select 9 as val from dual union all
select 10 as val from dual union all
select 13 as val from dual union all
select 15 as val from dual union all
select 16 as val from dual union all
select 17 as val from dual union all
select 18 as val from dual union all
select 19 as val from dual union all
select 23 as val from dual
),sq2 as(
select
row_number() over(order by 1) as rownumber,val
from stab
)
select
a.val,b.val
from sq2 A
join sq2 b on b.rownumber = a.rownumber+2
where mod(A.rownumber,3)=1
Output:
1 3
5 8
9 13
15 17
18 23
I want to aggregate the DAYS column based on the running distinct counts of CLIENT_ID, but the catch is CLIENT_ID that were seen from the previous DAYS should not be counted. How to do this in Oracle SQL?
Based on the table below (let's call this table DAY_CLIENT):
DAY CLIENT_ID
1 10
1 11
1 12
2 10
2 11
3 10
3 11
3 12
3 13
4 10
I want to get (let's call this table DAY_AGG):
DAYS CNT_CLIENT_ID
1 3
2 3
3 4
4 4
So, in day 1 there are 3 distinct client IDs.
In day 2, there are still 3 because CLIENT_ID 10 & 11 were already found in day 1. In day 3, distinct clients became 4 because CLIENT_ID 13 is not found on previous days.
Here's an alternative solution that may or may not be more performant than the other solutions:
WITH your_table AS (SELECT 1 DAY, 10 CLIENT_ID FROM dual UNION ALL
SELECT 1 DAY, 11 CLIENT_ID FROM dual UNION ALL
SELECT 1 DAY, 12 CLIENT_ID FROM dual UNION ALL
SELECT 2 DAY, 10 CLIENT_ID FROM dual UNION ALL
SELECT 2 DAY, 11 CLIENT_ID FROM dual UNION ALL
SELECT 3 DAY, 10 CLIENT_ID FROM dual UNION ALL
SELECT 3 DAY, 11 CLIENT_ID FROM dual UNION ALL
SELECT 3 DAY, 12 CLIENT_ID FROM dual UNION ALL
SELECT 3 DAY, 13 CLIENT_ID FROM dual UNION ALL
SELECT 4 DAY, 10 CLIENT_ID FROM dual)
SELECT DISTINCT DAY,
COUNT(CASE WHEN rn = 1 THEN client_id END) OVER (ORDER BY DAY) num_distinct_client_ids
FROM (SELECT DAY,
client_id,
row_number() OVER (PARTITION BY client_id ORDER BY DAY) rn
FROM your_table);
DAY NUM_DISTINCT_CLIENT_IDS
---------- -----------------------
1 3
2 3
3 4
4 4
I recommend you test all the solutions against your data to see which one works best for you.
One approach used a correlated subquery:
SELECT DISTINCT
d1.DAYS,
(SELECT COUNT(DISTINCT d2.CLIENT_ID) FROM yourTable d2
WHERE d2.DAYS <= d1.DAYS) AS CNT_CLIENT_ID
FROM yourTable d1
Here is a demo below for SQL Server, but it should also run on your Oracle. I always struggle with setting up Oracle demos.
Demo
You could also use apply operator if oracle support.
select day, CNT_CLIENT_ID
from DAY_CLIENT t cross apply (
select count(distinct CLIENT_ID) as CNT_CLIENT_ID
from DAY_CLIENT
where day <= t.day) tt
group by day, CNT_CLIENT_ID;
In other way use subquery with correlation approach
select day, (select count(distinct CLIENT_ID)
from DAY_CLIENT
where day <= t.day) as DAY_CLIENT
from DAY_CLIENT t
group by day;
Try to keep it simple, always. All other answers also good if you want to learn other ways. But in this case no need to be fancy at all.
SELECT days
, COUNT(DISTINCT client_id) cnt
FROM
(
SELECT 1 days, 10 client_id FROM dual --1
UNION ALL
SELECT 1, 11 FROM dual --2
UNION ALL
SELECT 1, 12 FROM dual --3
UNION ALL
SELECT 1, 11 FROM dual --4
UNION ALL
SELECT 2, 10 FROM dual
UNION ALL
SELECT 2, 11 FROM dual
UNION ALL
SELECT 2, 12 FROM dual
UNION ALL
SELECT 3, 10 FROM dual
UNION ALL
SELECT 3, 11 FROM dual
UNION ALL
SELECT 3, 12 FROM dual
UNION ALL
SELECT 3, 13 FROM dual
UNION ALL
SELECT 4, 10 FROM dual
)
GROUP BY days
ORDER BY 1
/
DAYS | CLIENT_ID
----------------
1 3
2 3
3 4
4 1