在 Oracle 11g 中将 1 列值显示到 3 个不同的列

Question

Hey guys i am usnig sql 11g. i came across one situation where my table have 1 column like this

column1
1
john
2000
2
david
4000
4
eon
5000

and i want output to split into 3 different columns like

id     name     salary
1      john      2000
2      david     3000
4      eon       5000

Answer 1

With just that 1 column in a table you will have dreadful trouble getting the desired output. In SQL tables are "unordered sets" so there is no guarantee that the data will be extracted in the exact same order you display it above - but to achieve the desired output we must have a reliable order of rows.

Assuming there is no other column in that table I would suggest you need to add an ID column like this:

alter table mytable
add (
    id NUMBER GENERATED BY DEFAULT AS IDENTITY
    )

+---------+----+
| COLUMN1 | ID |
+---------+----+
| 1       |  1 |
| john    |  2 |
| 2000    |  3 |
| 2       |  4 |
| david   |  5 |
| 4000    |  6 |
| 4       |  7 |
| eon     |  8 |
| 5000    |  9 |
+---------+----+

which (hopefully) will add the information we can then use to help order the output:

In the following we use lag by 1 row, and lead by 1 row to widen the data into 3 column, and then also use modulus of 3 on the id column to select just the wanted rows that have values in all 3 columns:

select id, name, salary
from (
    select 
      mod(id,3) AS mod3
    , lag(column1, 1) over(order by id)  as id
    , column1                            as name
    , lead(column1, 1) over(order by id) as salary
    from mytable
)
where mod3 = 2

+----+-------+--------+
| ID | NAME  | SALARY |
+----+-------+--------+
|  1 | john  |   2000 |
|  2 | david |   4000 |
|  4 | eon   |   5000 |
+----+-------+--------+

demo

Answer 2

Schema:

 create table table1 (column1 varchar(50));

 insert into table1
 with cte as (select '1'from dual
 union all
 select 'john' from dual
 union all
 select '2000'from dual
 union all
 select '2'from dual
 union all
 select 'david'from dual
 union all
 select '4000'from dual
 union all
 select '4'from dual
 union all
 select 'eon'from dual
 union all
 select '5000'from dual
 )
  select * from cte

Query:

 with cte as 
 (
     select column1,( ( Row_number()OVER(ORDER BY 1 ) - 1 ) / 3 ) + 1 rn from table1
 )
 , cte2 as
 (
     select column1,rn, row_number()over(partition by NVL(SUBSTR(rn, 0, INSTR(rn, '.')-1), rn) order by rn) rnk  ,NVL(SUBSTR(rn, 0, INSTR(rn, '.')-1), rn)grp
     from cte 
 )
 select max("1")ID,max("2")Name,max("3")Salary from(
 select *  from cte2)
 pivot (max(column1) for rnk in (1,2,3))
 group by grp
 order by grp
 

Output:

ID	NAME	SALARY
1	john	2000
2	david	4000
4	eon	5000

db<>fiddle here