0

我的数据看起来像：

MEDIA_ID | CHANNEL_NAME
EH/123A     CH-1
EH/123A     CH-4
EH/132A     CH-5
ES/133B     CH-1
ES/133B     CH-2
ES/133B     CH-5

我想要的是：

EH/123A  |  CH-1,CH-4,CH-5
ES/123B  |  CH-1,CH-2,CH-5

我在 Oracle 中使用这个 SQL：

SELECT DISTINCT 
PR.MEDIA_ID
, LISTAGG(PR.CHANNEL_NAME, ', ') WITHIN GROUP (ORDER BY CHANNEL_NAME) AS PREM_CHAN
FROM PREM_REPORT PR
GROUP BY PR.MEDIA_ITEM, PR.CHANNEL_NAME;

我得到的是：

MEDIA_ID | CHANNEL_NAME
EH/123A     CH-1,CH-1,CH-1,CH-1,CH-1,CH-1,CH-1,CH-1,CH-1,CH-1,CH-1,CH-1
EH/123A     CH-4,CH-4,CH-4,CH-4,CH-4,CH-4,CH-4,CH-4,CH-4,CH-4,CH-4,CH-4
EH/132A     CH-5,CH-5,CH-5,CH-5,CH-5,CH-5,CH-5,CH-5,CH-5,CH-5,CH-5,CH-5
ES/133B     CH-1,CH-1,CH-1,CH-1,CH-1,CH-1,CH-1,CH-1,CH-1,CH-1,CH-1,CH-1
ES/133B     CH-2,CH-2,CH-2,CH-2,CH-2,CH-2,CH-2,CH-2,CH-2,CH-2,CH-2,CH-2
ES/133B     CH-5,CH-5,CH-5,CH-5,CH-5,CH-5,CH-5,CH-5,CH-5,CH-5,CH-5,CH-5

想法？

谢谢。本

4

3 回答 3

1

我认为您想要的查询是：

SELECT PR.MEDIA_ID,
       LISTAGG(PR.CHANNEL_NAME, ', ') WITHIN GROUP (ORDER BY CHANNEL_NAME) AS PREM_CHAN
FROM PREM_REPORT PR
GROUP BY PR.MEDIA_ITEM;

也就是说，PR.CHANNEL_NAME从您的查询中删除。我不确定您为什么会通过您提供的查询获得结果。select distinct也许和之间存在一些奇怪的交互group by。你几乎从不使用select distinctwith group by。

编辑：

要在 中返回不同的值LIST_AGG()，您需要使用子查询。在这种情况下，一种简单的方法是：

SELECT PR.MEDIA_ID,
       LISTAGG(PR.CHANNEL_NAME, ', ') WITHIN GROUP (ORDER BY CHANNEL_NAME) AS PREM_CHAN
FROM (SELECT DISTINCT MEDIA_ID, CHANNEL_NAME
      FROM PREM_REPORT PR
     ) PR
GROUP BY PR.MEDIA_ITEM;
于 2015-11-17T15:27:17.553 回答
1

您可以删除GROUP BY并添加PARTITION BY：

SELECT DISTINCT PR.MEDIA_ID
   ,LISTAGG(PR.CHANNEL_NAME, ', ') 
   WITHIN GROUP (ORDER BY CHANNEL_NAME) OVER (PARTITION BY  PR.MEDIA_ID) AS PREM_CHAN
FROM PREM_REPORT PR;

SqlFiddleDemo

输出：

╔═══════════╦══════════════════╗
║ MEDIA_ID  ║    PREM_CHAN     ║
╠═══════════╬══════════════════╣
║ ES/133B   ║ CH-1, CH-2, CH-5 ║
║ EH/123A   ║ CH-1, CH-4, CH-5 ║
╚═══════════╩══════════════════╝
于 2015-11-17T15:33:12.983 回答
0

我认为你把 GROUP BY 弄错了。这对我有用。

WITH prem_report AS (
  SELECT 'EH/123A' media_id, 'CH-1' channel_name FROM DUAL
  UNION
  SELECT 'EH/123A' media_id, 'CH-4' channel_name FROM DUAL
  UNION
  SELECT 'EH/132A' media_id, 'CH-5' channel_name FROM DUAL
  UNION
  SELECT 'ES/133B' media_id, 'CH-1' channel_name FROM DUAL
  UNION
  SELECT 'ES/133B' media_id, 'CH-2' channel_name FROM DUAL
  UNION
  SELECT 'ES/133B' media_id, 'CH-5' channel_name FROM DUAL
)
SELECT DISTINCT pr.media_id, LISTAGG(pr.channel_name, ', ') WITHIN GROUP     (ORDER BY channel_name) AS prem_chan
FROM prem_report pr
GROUP BY pr.media_id
于 2015-11-17T15:32:25.707 回答