Oracle SQL 案例

Question

我的查询：

select school, rating, name
from mytable
where school = 'SchoolN';

结果：

+--------+------+------+ |学校|评分|姓名| +--------+------+------+ |学校N | A |姓名1| |学校N | B |姓名2| |学校N | A |姓名3| |学校N | A |姓名4| |学校N | A |姓名5| |学校N | B |姓名6| |学校N | B |姓名7| |学校N | B |姓名8| |学校N | B |姓名9| +--------+------+------+

我想要的是这样命名：

+--------+------+---------+ |学校|评分|姓名| +--------+------+---------+ |学校N |一个 |i-Name1 | |学校N | B |我的名字2 | |学校N | A |ii-名称3 | |学校N | A |iii-姓名4| |学校N | A |iv-Name5 | |学校N | B |ii-名称6 | |学校N | B |iii-姓名7| |学校N | B |iv-Name8 | |学校N | B |v-Name9 | +--------+------+---------+

我试过了：

select school, rating,
case when rownum = 1 then 'i-' || name
     when rownum = 2 then 'ii-' || name
     when rownum = 3 then 'iii-' || name
     when rownum = 4 then 'iv-' || name
     when rownum = 5 then 'v-' || name
end Name
from mytable
where school = 'SchoolN';

但是结果是错误的：

+--------+------+---------+ |学校|评分|姓名| +--------+------+---------+ |学校N |一个 |i-Name1 | |学校N | B |ii-名称2 | |学校N | A |iii-姓名3| |学校N | A |iv-Name4 | |学校N | A |v-Name5 | |学校N | B |姓名6 | |学校N | B |姓名7 | |学校N | B |姓名8 | |学校N | B |姓名9 | +--------+------+---------+

Answer 1

您可以在TO_CHAR 中使用罗马数字格式说明符RN

SELECT
    level as n,
    TO_CHAR(level,'RN') as roman
FROM
    dual
CONNECT BY
    level <= 10;


         N ROMAN          
---------- ---------------
         1               I
         2              II
         3             III
         4              IV
         5               V
         6              VI
         7             VII
         8            VIII
         9              IX
        10               X

10 rows selected. 

所以，在 APC 的查询中，您可以使用 to_char(rn,'RN') || name as name

Answer 2

ROWNUM 只是结果集中的行号。你真正想要的是每个rating 中name 的数量。所以这意味着你需要一个像row_number()这样的分析函数：

with cte as (
    select school, rating, name, 
           row_number() over (partition by  school, rating order by name) as rn       
    from mytable
)
select cte.school, 
       cte.rating,
       case when cte.rn = 1 then 'i-' || name
            when cte.rn = 2 then 'ii-' || name
            when cte.rn = 3 then 'iii-' || name
            when cte.rn = 4 then 'iv-' || name
            when cte.rn = 5 then 'v-' || name
        end Name
from cte
where cte.school = 'SchoolN'
;

这是生成所需结果集所需的对代码的最小更改。我建议你也合并@KaushikNayak 建议和deploy the rn format mask，因为这会让你的代码更紧凑，更不脆弱：

...
select cte.school, 
       cte.rating,
       to_char(cte.rn, 'rn')|| '-' || name as Name
from cte
...

传递小写格式掩码 ('rn') 以获得小写输出：i、ii 等。

Answer 3

你可以使用dense_rank()和fmRM格式模型选项的字符转换功能(to_char)：

select t.school,
       t.rating,
       to_char(dense_rank() over(partition by rating order by name), 'fmrm') 
       || '-' ||t.name name
  from mytable t
 order by t.name;

SQL Fiddle Demo