Oracle Group by salary range(0-999,1000-1999,2000-2999,...) [closed] - oracle

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user_id | username | salary |
+---------+----------+------+
| 1 | John | 4000 |
| 2 | Paul | 0900 |
| 3 | Adam | 0589 |
| 4 | Ben | 2154 |
| 5 | Charles | 2489 |
| 6 | Dean | 2500 |
| 7 | Edward | 2900 |
| 8 | Fred | 2800 |
| 9 | George | 4100 |
| 10 | Hugo | 5200 |
I need output like this
range count
--------------------
0-999 2
1000-1999 0
2000-2999 5
3000-3999 0
4000-4999 2
5000-5999 1

Answer 1

Here is an attempt:
with w as
(
select 1000 * (level - 1) low, 1000 * level high from dual
connect by level <= 10
)
select w.low, w.high, sum(decode(t.user_id, null, 0, 1)) nb
from w, test_epn t
where w.low <= t.salary (+)
and w.high > t.salary (+)
group by w.low, w.high
order by w.low
;
This gives:
1 0 1000 2
2 1000 2000 0
3 2000 3000 5
4 3000 4000 0
5 4000 5000 2
6 5000 6000 1
7 6000 7000 0
8 7000 8000 0
9 8000 9000 0
10 9000 10000 0

Answer 2

SQL> col range format a30
SQL> with t as (
2 select 'John' name, 4000 sal from dual union all
3 select 'Paul' name, 900 from dual union all
4 select 'Adam' name, 589 from dual union all
5 select 'Ben' name, 2154 from dual union all
6 select 'Charles' name, 2489 from dual union all
7 select 'Dean' name, 2500 from dual union all
8 select 'Edward' name, 2900 from dual union all
9 select 'Fred' name, 2800 from dual union all
10 select 'George' name, 4100 from dual union all
11 select 'Hugo' name, 5200 from dual
12 )
13 select to_char(pvtid*1000)||'-'||to_char(pvtid*1000+999) range, count(t.sal)
14 from t
15 ,
16 (
17 select rownum-1 pvtid
18 from dual connect by level <= (select floor(max(sal)/1000) from t)+1
19 ) piv
20 where piv.pvtid = floor(t.sal(+)/1000)
21 group by piv.pvtid
22 order by 1
23 /
RANGE COUNT(T.SAL)
------------------------------ ------------
0-999 2
1000-1999 0
2000-2999 5
3000-3999 0
4000-4999 2
5000-5999 1

Answer 3

Oracle 11g R2 Schema Setup:
create table test_table as
select 1 user_id, 'John' username , 4000 salary from dual union all
select 2 , 'Paul' , 0900 from dual union all
select 3 , 'Adam' , 0589 from dual union all
select 4 , 'Ben' , 2154 from dual union all
select 5 , 'Charles' , 2489 from dual union all
select 6 , 'Dean' , 2500 from dual union all
select 7 , 'Edward' , 2900 from dual union all
select 8 , 'Fred' , 2800 from dual union all
select 9 , 'George' , 4100 from dual union all
select 10 , 'Hugo' , 5200 from dual
Query 1:
with range_tab(f,t) as (select (level - 1)*1000 , (level - 1)*1000 + 999
from dual
connect by (level - 1)*1000 <= (select max(salary) from test_table))
select f ||'-'|| t as range, count(user_id)
from test_table
right outer join range_tab on (salary between f and t)
group by f, t
order by 1
[Results][2]:
| RANGE | COUNT(USER_ID) |
|-----------|----------------|
| 0-999 | 2 |
| 1000-1999 | 0 |
| 2000-2999 | 5 |
| 3000-3999 | 0 |
| 4000-4999 | 2 |
| 5000-5999 | 1 |

Answer 4

In case of fixed interval you can also use Oracle WIDTH_BUCKET function.
select count(*),
(WIDTH_BUCKET(salary, 0, 10000,10)-1)*1000 ||'-'||to_char(WIDTH_BUCKET(salary, 0, 10000,10)*1000-1) as salary_range
from table1
group by WIDTH_BUCKET(salary, 0, 10000,10)
order by salary_range;
| COUNT(*) | SALARY_RANGE |
|----------|--------------|
| 2 | 0-999 |
| 5 | 2000-2999 |
| 2 | 4000-4999 |
| 1 | 5000-5999 |
Disadvantage is: It does not count empty buckets, but maybe this satisfy your needs anyway.