oracle中RANK()和DENSE_RANK()函数之间有什么区别？

Question

RANK()和DENSE_RANK()功能有什么区别？如何找到下emptbl表中的第n个薪水？

DEPTNO  EMPNAME    SAL
------------------------------
10       rrr    10000.00
11       nnn    20000.00
11       mmm    5000.00
12       kkk    30000.00
10       fff    40000.00
10       ddd    40000.00
10       bbb    50000.00
10       ccc    50000.00

如果在表数据中有nulls,如果我想找nth工资会怎么样？

Answer 1

RANK为您提供订购分区内的排名.领带被分配相同的排名,跳过下一个排名.因此,如果您在排名2中有3个项目,则列出的下一个排名将排​​名为5.

DENSE_RANK再次为您提供有序分区内的排名,但排名是连续的.如果有多个项目的排名,则不会跳过任何排名.

至于null,它取决于ORDER BY子句.这是一个简单的测试脚本,您可以使用它来查看发生的情况:

with q as (
select 10 deptno, 'rrr' empname, 10000.00 sal from dual union all
select 11, 'nnn', 20000.00 from dual union all
select 11, 'mmm', 5000.00 from dual union all
select 12, 'kkk', 30000 from dual union all
select 10, 'fff', 40000 from dual union all
select 10, 'ddd', 40000 from dual union all
select 10, 'bbb', 50000 from dual union all
select 10, 'xxx', null from dual union all
select 10, 'ccc', 50000 from dual)
select empname, deptno, sal
     , rank() over (partition by deptno order by sal nulls first) r
     , dense_rank() over (partition by deptno order by sal nulls first) dr1
     , dense_rank() over (partition by deptno order by sal nulls last) dr2
 from q; 

EMP     DEPTNO        SAL          R        DR1        DR2
--- ---------- ---------- ---------- ---------- ----------
xxx         10                     1          1          4
rrr         10      10000          2          2          1
fff         10      40000          3          3          2
ddd         10      40000          3          3          2
ccc         10      50000          5          4          3
bbb         10      50000          5          4          3
mmm         11       5000          1          1          1
nnn         11      20000          2          2          2
kkk         12      30000          1          1          1

9 rows selected.

这是一个很好的解释和一些例子的链接.

Answer 2

这篇文章很好地解释了它.基本上,你可以这样看:

CREATE TABLE t AS
SELECT 'a' v FROM dual UNION ALL
SELECT 'a'   FROM dual UNION ALL
SELECT 'a'   FROM dual UNION ALL
SELECT 'b'   FROM dual UNION ALL
SELECT 'c'   FROM dual UNION ALL
SELECT 'c'   FROM dual UNION ALL
SELECT 'd'   FROM dual UNION ALL
SELECT 'e'   FROM dual;

SELECT
  v,
  ROW_NUMBER() OVER (ORDER BY v) row_number,
  RANK()       OVER (ORDER BY v) rank,
  DENSE_RANK() OVER (ORDER BY v) dense_rank
FROM t
ORDER BY v;

以上将产生:

+---+------------+------+------------+
| V | ROW_NUMBER | RANK | DENSE_RANK |
+---+------------+------+------------+
| a |          1 |    1 |          1 |
| a |          2 |    1 |          1 |
| a |          3 |    1 |          1 |
| b |          4 |    4 |          2 |
| c |          5 |    5 |          3 |
| c |          6 |    5 |          3 |
| d |          7 |    7 |          4 |
| e |          8 |    8 |          5 |
+---+------------+------+------------+

用语言说

• ROW_NUMBER() 为每一行赋予唯一值
• RANK() 将相同的行号归为相同的值,留下"漏洞"
• DENSE_RANK() 将相同的行号归为相同的值,不留任何"漏洞"

Answer 3

SELECT empno,
       deptno,
       sal,
       RANK() OVER (PARTITION BY deptno ORDER BY sal) "rank"
FROM   emp;

     EMPNO     DEPTNO        SAL       rank
---------- ---------- ---------- ----------
      7934         10       1300          1
      7782         10       2450          2
      7839         10       5000          3
      7369         20        800          1
      7876         20       1100          2
      7566         20       2975          3
      7788         20       3000          4
      7902         20       3000          4
      7900         30        950          1
      7654         30       1250          2
      7521         30       1250          2
      7844         30       1500          4
      7499         30       1600          5
      7698         30       2850          6


SELECT empno,
       deptno,
       sal,
       DENSE_RANK() OVER (PARTITION BY deptno ORDER BY sal) "rank"
FROM   emp;

     EMPNO     DEPTNO        SAL       rank
---------- ---------- ---------- ----------
      7934         10       1300          1
      7782         10       2450          2
      7839         10       5000          3
      7369         20        800          1
      7876         20       1100          2
      7566         20       2975          3
      7788         20       3000          4
      7902         20       3000          4
      7900         30        950          1
      7654         30       1250          2
      7521         30       1250          2
      7844         30       1500          3
      7499         30       1600          4
      7698         30       2850          5

Answer 4

select empno
       ,salary
       ,row_number() over(order by salary desc) as Serial
       ,Rank() over(order by salary desc) as rank
       ,dense_rank() over(order by salary desc) as denseRank
from emp ;

Row_number()-> 用于生成序列号

Dense_rank()将给出连续排名，但Rank()在排名冲突的情况下将跳过排名。

Answer 5

rank():用于对一组行中的记录进行排名.

dense_rank():DENSE_RANK函数的作用类似于RANK函数,只是它指定连续的行.

查询 -

select 
    ENAME,SAL,RANK() over (order by SAL) RANK
from 
    EMP;

输出 -

+--------+------+------+
| ENAME  | SAL  | RANK |
+--------+------+------+
| SMITH  |  800 |    1 |
| JAMES  |  950 |    2 |
| ADAMS  | 1100 |    3 |
| MARTIN | 1250 |    4 |
| WARD   | 1250 |    4 |
| TURNER | 1500 |    6 |
+--------+------+------+

查询 -

select 
    ENAME,SAL,dense_rank() over (order by SAL) DEN_RANK
from 
    EMP;

输出 -

+--------+------+-----------+
| ENAME  | SAL  |  DEN_RANK |
+--------+------+-----------+
| SMITH  |  800 |         1 |
| JAMES  |  950 |         2 |
| ADAMS  | 1100 |         3 |
| MARTIN | 1250 |         4 |
| WARD   | 1250 |         4 |
| TURNER | 1500 |         5 |
+--------+------+-----------+