Question

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 |

我需要这样的输出

range     count 
--------------------
 0-999      2

1000-1999   0

2000-2999   5

3000-3999   0

4000-4999   2

5000-5999   1

Answer 1

Oracle 11g R2架构设置：

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

查询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

<强> [结果] [2] ：

|     RANGE | COUNT(USER_ID) |
|-----------|----------------|
|     0-999 |              2 |
| 1000-1999 |              0 |
| 2000-2999 |              5 |
| 3000-3999 |              0 |
| 4000-4999 |              2 |
| 5000-5999 |              1 |

Answer 2

这是一次尝试：

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
;

这给出了：

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 3

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 4

如果是固定间隔，您还可以使用Oracle WIDTH_BUCKE T函数。

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 |

缺点是：它不计算空桶，但也许这可以满足您的需求。

Oracle Group按工资范围（0-999,1000-1999,2000-2999，...）

4 个答案: