甲骨文 ROW_NUMBER()

Question

我想写一个查询来显示这种结果

column0	column1	column2	column3
a	x	1	1
a	y	1	2
a	z	1	3
b	x	2	1
b	y	2	2
c	x	3	1
c	y	3	2
c	z	3	3

column2 -> column0 的行号

column3 -> 分区column0 & column1的行号

我已经试过了，但没有用

SELECT ROW_NUMBER() OVER (PARTITION BY column0 ORDER BY column0 ) column2,
       ROW_NUMBER() OVER (PARTITION BY column0 , column1  ORDER BY column0 , column1) column3  
FROM DUAL

你有什么想法吗？

Answer 1

对column2使用DENSE_RANK，对column3使用ROW_NUMBER:

SELECT
    column0,
    column1,
    DENSE_RANK() OVER (ORDER BY column0) AS column3,
    ROW_NUMBER() OVER (PARTITION BY column0 ORDER BY column1) AS column4
FROM yourTable;

Answer 2

与row_number:

SQL> with test (col0, col1) as
  2    (select 'a', 'x' from dual union all
  3     select 'a', 'y' from dual union all
  4     select 'a', 'z' from dual union all
  5     select 'b', 'x' from dual union all
  6     select 'b', 'y' from dual
  7    )
  8  select col0, col1,
  9    row_number() over (partition by col1 order by col0) col2,
 10    row_number() over (partition by col0 order by col1) col3
 11  from test
 12  order by col0, col1
 13  /

COL0  COL1         COL2       COL3
----- ------ ---------- ----------
a     x               1          1
a     y               1          2
a     z               1          3
b     x               2          1
b     y               2          2

SQL>

甲骨文 ROW_NUMBER()

oracle ROW_NUMBER()

oracle

row-number