Question

这个表与普通模式相反，我不知道如何从中获取所需的数据。

以下是一些示例数据

Value (column)          Info (column)
---------------------------------------------
Supplier_1              'Some supplier'
Supplier_1_email        'foo@gmail.com'
Supplier_1_rating       '5'
Supplier_1_status       'Active'
Supplier_2              'Some other supplier'
Supplier_2_email        'bar@gmail.com'
Supplier_2_rating       '4'
Supplier_2_status       'Active'
Supplier_3              'Yet another supplier'

...

我需要一个查询来查找评级最高且当前状态为“有效”的供应商的电子邮件。

Answer 1

select 
    m.sup_email, r.sup_rating 
from 
    (select substr(value, 1, length(value) - length('_email') as sup_name, info as sup_email from table where value like '%_email') as m 
left join 
    (select substr(value, 1, length(value) - length('_rating') as sup_name), info as sub_rating from table where value like '%_rating') as r on m.sup_name = r.sup_name 
order by 
   sup_rating desc 
limit 
    1;

Answer 2

对于单程解决方案，请尝试：

select "email" from 
(select 
    substr("value", 1, 8 + instr(substr("value", 10, length("value")-9),'_')) "supplier", 
    max(case when "value" like '%_status' then "info" end) as "status",
    max(case when "value" like '%_rating' then cast("info" as integer) end) as "rating",
    max(case when "value" like '%_email' then "info" end) as "email"
from "table" t
where "value" like '%_rating' or "value" like '%_email' or "value" like '%_status'
group by substr("value", 1, 8 + instr(substr("value", 10, length("value")-9),'_'))
having max(case when "value" like '%_status' then "info" end) = 'Active'
order by 3 desc
) where rownum = 1

（列名都是双引号，因为有些是保留字。）

Answer 3

扩展迈克的出色建议：

CREATE VIEW supplier_names AS
  SELECT SUBSTR(Value,INSTR(Value,'_')+1) AS supplier_id
        ,Info AS supplier_name
  FROM   the_table
  WHERE  INSTR(Value,'_',1,2) = 0;

CREATE VIEW supplier_emails AS
  SELECT SUBSTR(Value,INSTR(Value,'_')+1,INSTR(Value,'_',1,2)-INSTR(Value,'_')-1)
         AS supplier_id
        ,Info AS supplier_email
  FROM   the_table
  WHERE  Value LIKE '%email';

CREATE VIEW supplier_ratings AS
  SELECT SUBSTR(Value,INSTR(Value,'_')+1,INSTR(Value,'_',1,2)-INSTR(Value,'_')-1)
         AS supplier_id
        ,Info AS supplier_rating
  FROM   the_table
  WHERE  Value LIKE '%rating';

CREATE VIEW supplier_statuses AS
  SELECT SUBSTR(Value,INSTR(Value,'_')+1,INSTR(Value,'_',1,2)-INSTR(Value,'_')-1)
         AS supplier_id
        ,Info AS supplier_rating
  FROM   the_table
  WHERE  Value LIKE '%status';

查询将像狗一样执行，因此我建议您研究创建一些虚拟列，或至少基于函数的索引，以优化这些查询。

如何进行Oracle SQL查询？

3 个答案: