Question

oracle查询应该在单列中填充从单行到三个不同行的三个不同列值以及一行生成的一列并且结果不应在列中包含空值

表：电信
id ---提供者---消费者---接收者
1 ---- vodaphone ---- t-mobile ----- AT＆amp; T

预计出来是 id ----- name ---- type
1 ---- vodaphone ----提供者
1 ---- t-mobile ----消费者
1 ---- AT＆amp; T --------接收器

Answer 1

使用union all的简单方法：

select id, provider as name, 'provider' from telecom union all
select id, consumer, 'consumer' from telecom union all
select id, receiver, 'receiver' from telecom ;

这将扫描telecom三次。一种更有效的方法可能是：

select t.id,
       (case which when 'provider' then provider
                   when 'consumer' then consumer
                   when 'receiver' then receiver
        end) as name,
       which                  
from telecom t cross join
     (select 'provider' as which from dual union all
      select 'consumer' as which from dual union all
      select 'receiver' as which from dual
     ) x

Answer 2

Oracle安装程序：

SET DEFINE OFF;

CREATE TABLE Telecom ( id, provider, consumer, receiver ) AS
SELECT 1, 'vodaphone', 't-mobile', 'AT&T' FROM DUAL;

<强>查询：

SELECT id,
       name,
       "type"
FROM   telecom
UNPIVOT( name FOR "type" IN ( provider, consumer, receiver ) );

<强>输出：

       ID NAME      type   
---------- --------- --------
         1 vodaphone PROVIDER 
         1 t-mobile  CONSUMER 
         1 AT&T      RECEIVER

<强>更新：

SET DEFINE OFF;

CREATE TABLE Telecom ( id, col_provider, col_consumer, col_receiver ) AS
SELECT 1, 'vodaphone', 't-mobile', 'AT&T' FROM DUAL UNION ALL
SELECT 1, 't-mobile',  NULL, 'EE' FROM DUAL;

<强>查询：

SELECT id,
       name,
       "type"
FROM   telecom
UNPIVOT( name FOR "type" IN ( col_provider AS 'provider',
                              col_consumer AS 'consumer',
                              col_receiver AS 'receiver' ) );

<强>输出：

        ID NAME      type   
---------- --------- --------
         1 vodaphone provider 
         1 t-mobile  consumer 
         1 AT&T      receiver 
         1 t-mobile  provider 
         1 EE        receiver

oracle查询将不同的列填充到一列中

2 个答案: